Prove the following by using the principle of mathematical induction for all n∈N.
1. $1+3+3^{2}+...+3^{n-1}$ $=\frac{3^{n}-1}{2}$
Let P(n) be the statement $1+3+3^{2}+...+3^{n-1}$ $=\frac{3^{n}-1}{2}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1+3+3^{2}+...+3^{k-1}$ $=\frac{3^{k}-1}{2}$ ...(i)
P(k+1) ⇒ $1+3+3^{2}+...+3^{k-1}+3^{(k+1)-1}$ $=\frac{3^{k+1}-1}{2}$
L.H.S.,
= $1+3+3^{2}+...+3^{k-1}+3^{(k+1)-1}$
= $\frac{3^{k}-1}{2}+3^{k}$ [substituting (i)]
= $\frac{3^{k}-1+2*3^{k}}{2}$
= $\frac{3^{k}(1+2)-1}{2}$
= $\frac{3^{k}*3-1}{2}$
= $\frac{3^{k+1}-1}{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
2. $1^{3}+2^{3}+3^{3}+...+n^{3}$ $=\left ( \frac{n(n+1)}{2} \right )^{2}$
Let P(n) be the statement $1^{3}+2^{3}+3^{3}+...+n^{3}$ $=\left ( \frac{n(n+1)}{2} \right )^{2}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1^{3}+2^{3}+3^{3}+...+k^{3}$ $=\left ( \frac{k(k+1)}{2} \right )^{2}$
P(k+1) ⇒ $1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$ $=\left ( \frac{(k+1)(k+1+1)}{2} \right )^{2}$
$1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$ $=\left ( \frac{(k+1)(k+2)}{2} \right )^{2}$
L.H.S.,
= $1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$
= $\left ( \frac{k(k+1)}{2} \right )^{2}+(k+1)^{3}$
= $\frac{k^{2}}{4}(k+1)^{2}+(k+1)^{3}$
= $(k+1)^{2} \left [\frac{k^{2}}{4}+(k+1) \right ]$
= $(k+1)^{2} \left [\frac{k^{2}+4k+4}{4} \right ]$
= $(k+1)^{2} \left [\frac{(k+2)(k+2)}{4} \right ]$
= $(k+1)^{2} \left [\frac{(k+2)}{2} \right ]^{2}$
= $\left (\frac{(k+1)(k+2)}{2} \right )^{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
3. $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...n}$ $=\frac{2n}{n+1}$
Let P(n) be the statement $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...n}$ $=\frac{2n}{n+1}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $=\frac{2k}{k+1}$
P(k+1) ⇒ $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $+\frac{1}{1+2+3+...+(k+1)}$ $=\frac{2(k+1)}{(k+1)+1}$
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $+\frac{1}{1+2+3+...+(k+1)}$ $=\frac{2(k+1)}{(k+2)}$
L.H.S.,
= $\frac{2k}{k+1}$ $+\frac{1}{1+2+3+...+(k+1)}$
= $\frac{2k}{k+1}$ $+\frac{1}{\frac{(k+1)(k+2)}{2}}$
[$1+2+3+...n=\frac{n(n+1)}{2}$
$1+2+3+...+(k+1)$ $=\frac{(k+1)(k+2)}{2}$ ]
= $\frac{2k}{k+1}$ $+\frac{2}{(k+1)(k+2)}$
= $\frac{2k(k+2)+2}{(k+1)(k+2)}$
= $\frac{2(k^{2}+2k+1)}{(k+1)(k+2)}$
= $\frac{2(k+1)(k+1)}{(k+1)(k+2)}$
= $\frac{2(k+1)}{(k+2)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
4. $1.2.3+2.3.4+...$ $+n(n+1)(n+2)$ $=\frac{n(n+1)(n+2)(n+3)}{4}$
Let P(n) be the statement $1.2.3+2.3.4+...$ $+n(n+1)(n+2)$ $=\frac{n(n+1)(n+2)(n+3)}{4}$
P(1) ⇒ 6=6
P(1) is true
Let P(k) be true
⇒ $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $=\frac{k(k+1)(k+2)(k+3)}{4}$
P(k+1) ⇒ $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $+(k+1)((k+1)+1)((k+1)+2)$ $=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
L.H.S.,
= $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $+(k+1)((k+1)+1)((k+1)+2)$
= $\frac{k(k+1)(k+2)(k+3)}{4}$ $+(k+1)(k+2)(k+3)$
= $(k+1)(k+2)(k+3)$ $\left [ \frac{k}{4}+1 \right ]$
= $(k+1)(k+2)(k+3)$ $\left [ \frac{k+4}{4} \right ]$
= $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
5. $1.3+2.3^{2}+3.3^{3}+...+n.3^{n}$ $=\frac{(2n-1)3^{n+1}+3}{4}$
Let P(n) be the statement $1.3+2.3^{2}+3.3^{3}+...+n.3^{n}$ $=\frac{(2n-1)3^{n+1}+3}{4}$
P(1) ⇒ 3=3
P(1) is true
Let P(k) be true
⇒ $1.3+2.3^{2}+3.3^{3}+...+k.3^{k}$ $=\frac{(2k-1)3^{k+1}+3}{4}$
P(k+1) ⇒ $1.3+2.3^{2}+3.3^{3}+...$ $+k.3^{k}+(k+1).3^{k+1}$ $=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}$
$1.3+2.3^{2}+3.3^{3}+...$ $+k.3^{k}+(k+1).3^{k+1}$ $=\frac{(2k+1)3^{k+2}+3}{4}$
L.H.S.,
= $\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}$
= $\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$
= $\frac{3^{k+1}(2k-1+4k+4)+3}{4}$
= $\frac{3^{k+1}(6k+3)+3}{4}$
= $\frac{3^{k+1}.3(2k+1)+3}{4}$
= $\frac{(2k+1)3^{k+2}+3}{4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
6. $1.2+2.3+3.4+...$ $+n.(n+1)$ $=\frac{n(n+1)(n+2)}{3}$
Let P(n) be the statement $1.2+2.3+3.4+...$ $+n.(n+1)$ $=\frac{n(n+1)(n+2)}{3}$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $1.2+2.3+3.4+...$ $+k.(k+1)$ $=\frac{k(k+1)(k+2)}{3}$
P(k+1) ⇒ $1.2+2.3+3.4+...$ $+k.(k+1)$ $+(k+1).((k+1)+1)$ $=\frac{(k+1)(k+2)(k+3)}{3}$
L.H.S.,
= $1.2+2.3+3.4+...$ $+k.(k+1)$$+(k+1).(k+2)$
= $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$
= $(k+1)(k+2)\left [ \frac{k}{3}+1 \right ]$
= $(k+1)(k+2)\left [ \frac{k+3}{3} \right ]$
= $\frac{(k+1)(k+2)(k+3)}{3}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
7. $1.3+3.5+5.7+...$ $+(2n-1)(2n+1)$ $= \frac{n(4n^{2}+n-1)}{3}$
Let P(n) be the statement $1.3+3.5+5.7+...$ $+(2n-1)(2n+1)$ $= \frac{n(4n^{2}+6n-1)}{3}$
P(1) ⇒ 3=3
P(1) is true
Let P(k) be true
⇒ $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $= \frac{k(4k^{2}+6k-1)}{3}$
P(k+1) ⇒ $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $+(2(k+1)-1)(2(k+1)+1)$ $= \frac{(k+1)(4(k+1)^{2}+6(k+1)-1)}{3}$
L.H.S.,
= $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $+(2(k+1)-1)(2(k+1)+1)$
= $ \frac{k(4k^{2}+6k-1)}{3}$ $+(2k+1)(2k+3)$
= $ \frac{k(4k^{2}+6k-1)}{3}$ $+(4k^{2}+8k+3)$
= $\frac{4k^{3}+6k^{2}-k+3(4k^{2}+8k+3)}{3}$
= $\frac{1}{3}\left [ 4k^{3}+18k^{2}+23k+9 \right ]$
= $\frac{1}{3}\left [(k+1)(4k^{2}+14k+9) \right ]$
R.H.S.,
= $\small \frac{1}{3}\left [(k+1)(4(k+1)^{2}+6(k+1)-1) \right ]$
= $\frac{1}{3}\left [(k+1)(4k^{2}+14k+9) \right ]$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
8. $1.2+2.2^{2}+3.2^{3}+...$ $+n.2^{n}$ $=(n-1)2^{n+1}+2$
Let P(n) be the statement $1.2+2.2^{2}+3.2^{3}+...$ $+n.2^{n}$ $=(n-1)2^{n+1}+2$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $=(k-1)2^{k+1}+2$
P(k+1) ⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $+(k+1).2^{k+1}$ $=((k+1)-1)2^{k+1+1}+2$
⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $+(k+1).2^{k+1}$ $=k.2^{k+2}+2$
L.H.S.,
= $(k-1)2^{k+1}+2$ $+(k+1).2^{k+1}$
= $2^{k+1}\left ( k-1+k+1 \right )+2$
= $2^{k+1}\left ( 2k \right )+2$
= $k.2^{k+2}+2$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
9. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{n}}$ $=1-\frac{1}{2^{n}}$
Let P(n) be the statement $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{n}}$ $=1-\frac{1}{2^{n}}$
P(1) ⇒ 1/2=1/2
P(1) is true
Let P(k) be true
⇒ $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $=1-\frac{1}{2^{k}}$
P(k+1) ⇒ $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$ $=1-\frac{1}{2^{k+1}}$
L.H.S.,
= $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$
= $1-\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$
= $1-\frac{1}{2^{k}}+\frac{1}{2^{k}.2}$
= $1-\frac{1}{2^{k}}(1-\frac{1}{2})$
= $1-\frac{1}{2^{k}}(\frac{1}{2})$
= $1-\frac{1}{2^{k+1}}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
10. $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3n-1)(3n+2)}$ $=\frac{n}{6n+4}$
Let P(n) be the statement $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3n-1)(3n+2)}$ $=\frac{n}{6n+4}$
P(1) ⇒ 1/10=1/10
P(1) is true
Let P(k) be true
⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $=\frac{k}{6k+4}$
P(k+1) ⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$ $=\frac{k+1}{6(k+1)+4}$
⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$ $=\frac{k+1}{6k+10}$
L.H.S.,
= $\frac{k}{2(3k+2)}$ $+\frac{1}{(3k+2)(3k+5)}$
= $\frac{k(3k+5)+2}{2(3k+2)(3k+5)}$
= $\frac{3k^{2}+5k+2}{2(3k+2)(3k+5)}$
= $\frac{(3k+2)(k+1)}{2(3k+2)(3k+5)}$
= $\frac{k+1}{2(3k+5)}$
= $\frac{k+1}{6k+10}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
11. $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{n(n+1)(n+2)}$ $=\frac{n(n+3)}{4(n+1)(n+2)}$
Let P(n) be the statement $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{n(n+1)(n+2)}$ $=\frac{n(n+3)}{4(n+1)(n+2)}$
P(1) ⇒ 1/6=1/6
P(1) is true
Let P(k) be true
⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $=\frac{k(k+3)}{4(k+1)(k+2)}$
P(k+1) ⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $+\frac{1}{(k+1)((k+1)+1)((k+1)+2)}$ $=\frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)}$
⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $+\frac{1}{(k+1)(k+2)(k+3)}$ $=\frac{(k+1)(k+4)}{4(k+2)(k+3)}$
L.H.S.,
= $\frac{k(k+3)}{4(k+1)(k+2)}$ $+\frac{1}{(k+1)(k+2)(k+3)}$
= $\frac{k(k+3)(k+3)+4}{4(k+1)(k+2)(k+3)}$
= $\frac{k^{3}+6k^{2}+9k+4}{4(k+1)(k+2)(k+3)}$
= $\frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}$
= $\frac{(k+1)(k+4)}{4(k+2)(k+3)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
12. $a+ar+ar^{2}+...$ $+ar^{n-1}$ $=\frac{a(r^{n}-1)}{r-1}$
Let P(n) be the statement $a+ar+ar^{2}+...$ $+ar^{n-1}$ $=\frac{a(r^{n}-1)}{r-1}$
P(1) ⇒ a=a
P(1) is true
Let P(k) be true
⇒ $a+ar+ar^{2}+...$ $+ar^{k-1}$ $=\frac{a(r^{k}-1)}{r-1}$
P(k+1) ⇒ $a+ar+ar^{2}+...$ $+ar^{k-1}$ $+ar^{k+1-1}$ $=\frac{a(r^{k+1}-1)}{r-1}$
L.H.S.,
= $a+ar+ar^{2}+...$ $+ar^{k-1}$ $+ar^{k}$
= $\frac{a(r^{k}-1)}{r-1}$ $+ar^{k}$
= $\frac{ar^{k}-a+(r-1)ar^{k}}{r-1}$
= $\frac{ar^{k}(1+r-1)-a}{r-1}$
= $\frac{ar^{k}(r)-a}{r-1}$
= $\frac{a(r^{k+1}-1)}{r-1}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
13. $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2n+1}{n^{2}} \right )$ $=(n+1)^{2}$
Let P(n) be the statement $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2n+1}{n^{2}} \right )$ $=(n+1)^{2}$
P(1) ⇒ 4=4
P(1) is true
Let P(k) be true
⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $=(k+1)^{2}$
P(k+1) ⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $\left ( 1+\frac{2(k+1)+1}{(k+1)^{2}} \right )$ $=(k+1+1)^{2}$
⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $\left ( 1+\frac{2k+3}{(k+1)^{2}} \right )$ $=(k+2)^{2}$
L.H.S.,
= $(k+1)^{2}$ $\left ( 1+\frac{2k+3}{(k+1)^{2}} \right )$
= $(k+1)^{2}$ $\frac{(k+1)^{2}+2k+3}{(k+1)^{2}}$
= $k^{2}+4k+4$
= $(k+2)(k+2)$
= $(k+2)^{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
14. $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{n} \right )$ $=n+1$
Let P(n) be the statement $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{n} \right )$ $=n+1$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $=k+1$
P(k+1) ⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $\left ( 1+\frac{1}{k+1} \right )$ $=k+1+1$
⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $\left ( 1+\frac{1}{k+1} \right )$ $=k+2$
L.H.S.,
= $(k+1)$ $\left ( 1+\frac{1}{k+1} \right )$
= $(k+1)$ $\left ( \frac{k+1+1}{k+1} \right )$
= k+2
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
15. $1^{2}+3^{2}+5^{2}+...$ $+(2n-1)^{2}$ $=\frac{n(2n-1)(2n+1)}{3}$
Let P(n) be the statement $1^{2}+3^{2}+5^{2}+...$ $+(2n-1)^{2}$ $=\frac{n(2n-1)(2n+1)}{3}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $=\frac{k(2k-1)(2k+1)}{3}$
P(k+1) ⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $+(2(k+1)-1)^{2}$ $=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$
⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $+(2k+1)^{2}$ $=\frac{(k+1)(2k+1)(2k+3)}{3}$
L.H.S.,
= $\frac{k(2k-1)(2k+1)}{3}$ $+(2k+1)^{2}$
= $\frac{k(2k-1)(2k+1)+3(2k+1)^{2}}{3}$
= $\frac{(2k+1)}{3}[k(2k-1)+3(2k+1)]$
= $\frac{(2k+1)}{3}[2k^{2}+5k+3]$
= $\frac{(2k+1)}{3}[(2k+3)(k+1)]$
= $\frac{(k+1)(2k+1)(2k+3)}{3}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
16. $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3n-2)(3n+1)}$ $=\frac{n}{3n+1}$
Let P(n) be the statement $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3n-2)(3n+1)}$ $=\frac{n}{3n+1}$
P(1) ⇒ 1/4=1/4
P(1) is true
Let P(k) be true
⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $=\frac{k}{3k+1}$
P(k+1) ⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$ $=\frac{k+1}{3(k+1)+1}$
⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $+\frac{1}{(3k+1)(3k+4)}$ $=\frac{k+1}{3k+4}$
L.H.S.,
= $\frac{k}{3k+1}$ $+\frac{1}{(3k+1)(3k+4)}$
= $\frac{k(3k+4)+1}{(3k+1)(3k+4)}$
= $\frac{3k^{2}+4k+1}{(3k+1)(3k+4)}$
= $\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
= $\frac{k+1}{3k+4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
17. $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2n+1)(2n+3)}$ $=\frac{n}{3(2n+3)}$
Let P(n) be the statement $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2n+1)(2n+3)}$ $=\frac{n}{3(2n+3)}$
P(1) ⇒ 1/15=1/15
P(1) is true
Let P(k) be true
⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $=\frac{k}{3(2k+3)}$
P(k+1) ⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$ $=\frac{k+1}{3(2(k+1)+3)}$
⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $+\frac{1}{(2k+3)(2k+5)}$ $=\frac{k+1}{3(2k+5)}$
L.H.S.,
= $\frac{k}{3(2k+3)}$ $+\frac{1}{(2k+3)(2k+5)}$
= $\frac{k(2k+5)+3}{3(2k+3)(2k+5)}$
= $\frac{2k^{2}+5k+3}{3(2k+3)(2k+5)}$
= $\frac{(2k+3)(k+1)}{3(2k+3)(2k+5)}$
= $\frac{k+1}{3(2k+5)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
18. $1+2+3+...+n$ $< \frac{1}{8}(2n+1)^{2}$
Let P(n) be the statement $1+2+3+...+n$ $< \frac{1}{8}(2n+1)^{2}$
P(1) ⇒ $1< \frac{9}{8}$
P(1) is true
Let P(k) be true
⇒ $1+2+3+...+k$ $< \frac{1}{8}(2k+1)^{2}$
P(k+1) ⇒ $1+2+3+...+k+(k+1)$ $< \frac{1}{8}(2(k+1)+1)^{2}$
⇒ $1+2+3+...+k+(k+1)$ $< \frac{1}{8}(2k+3)^{2}$
L.H.S.,
$< \frac{1}{8}(2k+1)^{2}+k+1$
$< \frac{1}{8}(4k^{2}+1+4k+8k+8)$
$< \frac{1}{8}(4k^{2}+12k+9)$
$< \frac{1}{8}(2k+3)(2k+3)$
$< \frac{1}{8}(2k+3)^{2}$
L.H.S. $< \frac{1}{8}(2k+3)^{2}$
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
19. n(n+1)(n+5) is a multiple of 3.
Let P(n) be the statement n(n+1)(n+5)
Let P(k) be true
⇒ k(k+1)(k+5)=3q, for some integer q.
P(k+1) ⇒ (k+1) (k+2) (k+5+1)
= (k+1) (k+2) ((k+5)+1)
= (k+1) (k+2) (k+5)+ (k+1) (k+2)
= k (k+1) (k+5)+2 (k+1) (k+5)+ (k+1) (k+2)
= 3q+2 (k+1) (k+5)+ (k+1) (k+2)
= 3q+ 3k2+15k+12
= 3q+3(k+1) (k+4)
= 3[ q+ (k+1) (k+4)], which is multiple of 3.
⇒ P(k+1) is true
20. $10^{2n-1}+1$ is divisible by 11
Let P(n) be the statement $10^{2n-1}+1$
Let P(k) be true
⇒ $10^{2k-1}+1=11q$ , for some integer q.
⇒ $10^{2k-1}=11q-1$
P(k+1) ⇒ $10^{2(k+1)-1}+1$
= $10^{2k+2-1}+1$
= $10^{2k-1}.10^{2}+1$
= $(11q-1).100+1$
= $11q.100-100+1$
= $11q.100-99$
= $11(100q-9)$ , which is divisible by 11.
⇒ P(k+1) is true
21. $x^{2n}-y^{2n}$ is divisible by x+y
Let P(n) be the statement $x^{2n}-y^{2n}$
Let P(k) be true
⇒ $x^{2k}-y^{2k}=(x+y)f(x,y)$
⇒ $x^{2n}=(x+y)f(x,y)+y^{2n}$
P(k+1) ⇒ $x^{2(k+1)}-y^{2(k+1)}$
= $x^{2k}x^{2}-y^{2k}y^{2}$
= $[(x+y)f(x,y)+y^{2k}]x^{2}-y^{2k}y^{2}$
= $(x+y)f(x,y)x^{2}+y^{2k}x^{2}-y^{2k}y^{2}$
= $(x+y)f(x,y)x^{2}+y^{2k}(x^{2}-y^{2})$
= $(x+y)f(x,y)x^{2}+y^{2k}(x-y)(x+y)$
= $(x+y)\left [f(x,y)x^{2}+y^{2k}(x-y) \right ]$ , which is divisible by (x+y)
⇒ P(k+1) is true
22. $3^{2n+2}-8n-9$ is divisible by 8.
Let P(n) be the statement $3^{2n+2}-8n-9$
Let P(k) be true
⇒ $3^{2k+2}-8k-9=8q$
⇒ $3^{2k+2}=8q+8k+9$
P(k+1) ⇒ $3^{2(k+1)+2}-8(k+1)-9$
= $3^{2k+2}.3^{2}-8k-8-9$
= $[8q+8k+9].9-8k-17$
= $8q.9+8k.9+81-8k-17$
= $8q.9+8k.8+64$
= $8[9q+8k+8]$ , which is divisible by 8
⇒ P(k+1) is true
23. $41^{n}-14^{n}$ is multiple of 27
Let P(n) be the statement $41^{n}-14^{n}$
Let P(k) be true
⇒ $41^{k}-14^{k}=27q$
⇒ $41^{k}=27q+14^{k}$
P(k+1) ⇒ $41^{k+1}-14^{k+1}$
= $41^{k}.41-14^{k}.14$
= $[27q+14^{k}].41-14^{k}.14$
= $27q.41+14^{k}.41-14^{k}.14$
= $27q.41+14^{k}.27$
= $27[41q+14^{k}]$ , which is multiple of 27
⇒ P(k+1) is true
24. $(2n+7)< (n+3)^{2}$
Let P(n) be the statement $(2n+7)< (n+3)^{2}$
P(1) ⇒ 9<16
P(1) is true
Let P(k) be true
⇒ $(2k+7)< (k+3)^{2}$
P(k+1) ⇒ $(2(k+1)+7)< ((k+1)+3)^{2}$
⇒ $(2k+7)+2< (k+4)^{2}$
L.H.S.,
$<(k+3)^{2}+2$
$<k^{2}+6k+11$
Now, $k^{2}+6k+11$ $<k^{2}+8k+16$
∴ $(2(k+1)+7)< (k+4)^{2}$
⇒ P(k+1) is true
1. $1+3+3^{2}+...+3^{n-1}$ $=\frac{3^{n}-1}{2}$
Let P(n) be the statement $1+3+3^{2}+...+3^{n-1}$ $=\frac{3^{n}-1}{2}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1+3+3^{2}+...+3^{k-1}$ $=\frac{3^{k}-1}{2}$ ...(i)
P(k+1) ⇒ $1+3+3^{2}+...+3^{k-1}+3^{(k+1)-1}$ $=\frac{3^{k+1}-1}{2}$
L.H.S.,
= $1+3+3^{2}+...+3^{k-1}+3^{(k+1)-1}$
= $\frac{3^{k}-1}{2}+3^{k}$ [substituting (i)]
= $\frac{3^{k}-1+2*3^{k}}{2}$
= $\frac{3^{k}(1+2)-1}{2}$
= $\frac{3^{k}*3-1}{2}$
= $\frac{3^{k+1}-1}{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
2. $1^{3}+2^{3}+3^{3}+...+n^{3}$ $=\left ( \frac{n(n+1)}{2} \right )^{2}$
Let P(n) be the statement $1^{3}+2^{3}+3^{3}+...+n^{3}$ $=\left ( \frac{n(n+1)}{2} \right )^{2}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1^{3}+2^{3}+3^{3}+...+k^{3}$ $=\left ( \frac{k(k+1)}{2} \right )^{2}$
P(k+1) ⇒ $1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$ $=\left ( \frac{(k+1)(k+1+1)}{2} \right )^{2}$
$1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$ $=\left ( \frac{(k+1)(k+2)}{2} \right )^{2}$
L.H.S.,
= $1^{3}+2^{3}+3^{3}+...+k^{3}+(k+1)^{3}$
= $\left ( \frac{k(k+1)}{2} \right )^{2}+(k+1)^{3}$
= $\frac{k^{2}}{4}(k+1)^{2}+(k+1)^{3}$
= $(k+1)^{2} \left [\frac{k^{2}}{4}+(k+1) \right ]$
= $(k+1)^{2} \left [\frac{k^{2}+4k+4}{4} \right ]$
= $(k+1)^{2} \left [\frac{(k+2)(k+2)}{4} \right ]$
= $(k+1)^{2} \left [\frac{(k+2)}{2} \right ]^{2}$
= $\left (\frac{(k+1)(k+2)}{2} \right )^{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
3. $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...n}$ $=\frac{2n}{n+1}$
Let P(n) be the statement $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...n}$ $=\frac{2n}{n+1}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $=\frac{2k}{k+1}$
P(k+1) ⇒ $1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $+\frac{1}{1+2+3+...+(k+1)}$ $=\frac{2(k+1)}{(k+1)+1}$
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ $+\frac{1}{1+2+3+...k}$ $+\frac{1}{1+2+3+...+(k+1)}$ $=\frac{2(k+1)}{(k+2)}$
L.H.S.,
= $\frac{2k}{k+1}$ $+\frac{1}{1+2+3+...+(k+1)}$
= $\frac{2k}{k+1}$ $+\frac{1}{\frac{(k+1)(k+2)}{2}}$
[$1+2+3+...n=\frac{n(n+1)}{2}$
$1+2+3+...+(k+1)$ $=\frac{(k+1)(k+2)}{2}$ ]
= $\frac{2k}{k+1}$ $+\frac{2}{(k+1)(k+2)}$
= $\frac{2k(k+2)+2}{(k+1)(k+2)}$
= $\frac{2(k^{2}+2k+1)}{(k+1)(k+2)}$
= $\frac{2(k+1)(k+1)}{(k+1)(k+2)}$
= $\frac{2(k+1)}{(k+2)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
4. $1.2.3+2.3.4+...$ $+n(n+1)(n+2)$ $=\frac{n(n+1)(n+2)(n+3)}{4}$
Let P(n) be the statement $1.2.3+2.3.4+...$ $+n(n+1)(n+2)$ $=\frac{n(n+1)(n+2)(n+3)}{4}$
P(1) ⇒ 6=6
P(1) is true
Let P(k) be true
⇒ $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $=\frac{k(k+1)(k+2)(k+3)}{4}$
P(k+1) ⇒ $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $+(k+1)((k+1)+1)((k+1)+2)$ $=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
L.H.S.,
= $1.2.3+2.3.4+...$ $+k(k+1)(k+2)$ $+(k+1)((k+1)+1)((k+1)+2)$
= $\frac{k(k+1)(k+2)(k+3)}{4}$ $+(k+1)(k+2)(k+3)$
= $(k+1)(k+2)(k+3)$ $\left [ \frac{k}{4}+1 \right ]$
= $(k+1)(k+2)(k+3)$ $\left [ \frac{k+4}{4} \right ]$
= $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
5. $1.3+2.3^{2}+3.3^{3}+...+n.3^{n}$ $=\frac{(2n-1)3^{n+1}+3}{4}$
Let P(n) be the statement $1.3+2.3^{2}+3.3^{3}+...+n.3^{n}$ $=\frac{(2n-1)3^{n+1}+3}{4}$
P(1) ⇒ 3=3
P(1) is true
Let P(k) be true
⇒ $1.3+2.3^{2}+3.3^{3}+...+k.3^{k}$ $=\frac{(2k-1)3^{k+1}+3}{4}$
P(k+1) ⇒ $1.3+2.3^{2}+3.3^{3}+...$ $+k.3^{k}+(k+1).3^{k+1}$ $=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}$
$1.3+2.3^{2}+3.3^{3}+...$ $+k.3^{k}+(k+1).3^{k+1}$ $=\frac{(2k+1)3^{k+2}+3}{4}$
L.H.S.,
= $\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}$
= $\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$
= $\frac{3^{k+1}(2k-1+4k+4)+3}{4}$
= $\frac{3^{k+1}(6k+3)+3}{4}$
= $\frac{3^{k+1}.3(2k+1)+3}{4}$
= $\frac{(2k+1)3^{k+2}+3}{4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
6. $1.2+2.3+3.4+...$ $+n.(n+1)$ $=\frac{n(n+1)(n+2)}{3}$
Let P(n) be the statement $1.2+2.3+3.4+...$ $+n.(n+1)$ $=\frac{n(n+1)(n+2)}{3}$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $1.2+2.3+3.4+...$ $+k.(k+1)$ $=\frac{k(k+1)(k+2)}{3}$
P(k+1) ⇒ $1.2+2.3+3.4+...$ $+k.(k+1)$ $+(k+1).((k+1)+1)$ $=\frac{(k+1)(k+2)(k+3)}{3}$
L.H.S.,
= $1.2+2.3+3.4+...$ $+k.(k+1)$$+(k+1).(k+2)$
= $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$
= $(k+1)(k+2)\left [ \frac{k}{3}+1 \right ]$
= $(k+1)(k+2)\left [ \frac{k+3}{3} \right ]$
= $\frac{(k+1)(k+2)(k+3)}{3}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
7. $1.3+3.5+5.7+...$ $+(2n-1)(2n+1)$ $= \frac{n(4n^{2}+n-1)}{3}$
Let P(n) be the statement $1.3+3.5+5.7+...$ $+(2n-1)(2n+1)$ $= \frac{n(4n^{2}+6n-1)}{3}$
P(1) ⇒ 3=3
P(1) is true
Let P(k) be true
⇒ $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $= \frac{k(4k^{2}+6k-1)}{3}$
P(k+1) ⇒ $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $+(2(k+1)-1)(2(k+1)+1)$ $= \frac{(k+1)(4(k+1)^{2}+6(k+1)-1)}{3}$
L.H.S.,
= $1.3+3.5+5.7+...$ $+(2k-1)(2k+1)$ $+(2(k+1)-1)(2(k+1)+1)$
= $ \frac{k(4k^{2}+6k-1)}{3}$ $+(2k+1)(2k+3)$
= $ \frac{k(4k^{2}+6k-1)}{3}$ $+(4k^{2}+8k+3)$
= $\frac{4k^{3}+6k^{2}-k+3(4k^{2}+8k+3)}{3}$
= $\frac{1}{3}\left [ 4k^{3}+18k^{2}+23k+9 \right ]$
= $\frac{1}{3}\left [(k+1)(4k^{2}+14k+9) \right ]$
R.H.S.,
= $\small \frac{1}{3}\left [(k+1)(4(k+1)^{2}+6(k+1)-1) \right ]$
= $\frac{1}{3}\left [(k+1)(4k^{2}+14k+9) \right ]$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
8. $1.2+2.2^{2}+3.2^{3}+...$ $+n.2^{n}$ $=(n-1)2^{n+1}+2$
Let P(n) be the statement $1.2+2.2^{2}+3.2^{3}+...$ $+n.2^{n}$ $=(n-1)2^{n+1}+2$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $=(k-1)2^{k+1}+2$
P(k+1) ⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $+(k+1).2^{k+1}$ $=((k+1)-1)2^{k+1+1}+2$
⇒ $1.2+2.2^{2}+3.2^{3}+...$ $+k.2^{k}$ $+(k+1).2^{k+1}$ $=k.2^{k+2}+2$
L.H.S.,
= $(k-1)2^{k+1}+2$ $+(k+1).2^{k+1}$
= $2^{k+1}\left ( k-1+k+1 \right )+2$
= $2^{k+1}\left ( 2k \right )+2$
= $k.2^{k+2}+2$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
9. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{n}}$ $=1-\frac{1}{2^{n}}$
Let P(n) be the statement $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{n}}$ $=1-\frac{1}{2^{n}}$
P(1) ⇒ 1/2=1/2
P(1) is true
Let P(k) be true
⇒ $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $=1-\frac{1}{2^{k}}$
P(k+1) ⇒ $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$ $=1-\frac{1}{2^{k+1}}$
L.H.S.,
= $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ $+\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$
= $1-\frac{1}{2^{k}}$ $+\frac{1}{2^{k+1}}$
= $1-\frac{1}{2^{k}}+\frac{1}{2^{k}.2}$
= $1-\frac{1}{2^{k}}(1-\frac{1}{2})$
= $1-\frac{1}{2^{k}}(\frac{1}{2})$
= $1-\frac{1}{2^{k+1}}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
10. $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3n-1)(3n+2)}$ $=\frac{n}{6n+4}$
Let P(n) be the statement $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3n-1)(3n+2)}$ $=\frac{n}{6n+4}$
P(1) ⇒ 1/10=1/10
P(1) is true
Let P(k) be true
⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $=\frac{k}{6k+4}$
P(k+1) ⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$ $=\frac{k+1}{6(k+1)+4}$
⇒ $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...$ $+\frac{1}{(3k-1)(3k+2)}$ $+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$ $=\frac{k+1}{6k+10}$
L.H.S.,
= $\frac{k}{2(3k+2)}$ $+\frac{1}{(3k+2)(3k+5)}$
= $\frac{k(3k+5)+2}{2(3k+2)(3k+5)}$
= $\frac{3k^{2}+5k+2}{2(3k+2)(3k+5)}$
= $\frac{(3k+2)(k+1)}{2(3k+2)(3k+5)}$
= $\frac{k+1}{2(3k+5)}$
= $\frac{k+1}{6k+10}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
11. $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{n(n+1)(n+2)}$ $=\frac{n(n+3)}{4(n+1)(n+2)}$
Let P(n) be the statement $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{n(n+1)(n+2)}$ $=\frac{n(n+3)}{4(n+1)(n+2)}$
P(1) ⇒ 1/6=1/6
P(1) is true
Let P(k) be true
⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $=\frac{k(k+3)}{4(k+1)(k+2)}$
P(k+1) ⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $+\frac{1}{(k+1)((k+1)+1)((k+1)+2)}$ $=\frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)}$
⇒ $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...$ $+\frac{1}{k(k+1)(k+2)}$ $+\frac{1}{(k+1)(k+2)(k+3)}$ $=\frac{(k+1)(k+4)}{4(k+2)(k+3)}$
L.H.S.,
= $\frac{k(k+3)}{4(k+1)(k+2)}$ $+\frac{1}{(k+1)(k+2)(k+3)}$
= $\frac{k(k+3)(k+3)+4}{4(k+1)(k+2)(k+3)}$
= $\frac{k^{3}+6k^{2}+9k+4}{4(k+1)(k+2)(k+3)}$
= $\frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}$
= $\frac{(k+1)(k+4)}{4(k+2)(k+3)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
12. $a+ar+ar^{2}+...$ $+ar^{n-1}$ $=\frac{a(r^{n}-1)}{r-1}$
Let P(n) be the statement $a+ar+ar^{2}+...$ $+ar^{n-1}$ $=\frac{a(r^{n}-1)}{r-1}$
P(1) ⇒ a=a
P(1) is true
Let P(k) be true
⇒ $a+ar+ar^{2}+...$ $+ar^{k-1}$ $=\frac{a(r^{k}-1)}{r-1}$
P(k+1) ⇒ $a+ar+ar^{2}+...$ $+ar^{k-1}$ $+ar^{k+1-1}$ $=\frac{a(r^{k+1}-1)}{r-1}$
L.H.S.,
= $a+ar+ar^{2}+...$ $+ar^{k-1}$ $+ar^{k}$
= $\frac{a(r^{k}-1)}{r-1}$ $+ar^{k}$
= $\frac{ar^{k}-a+(r-1)ar^{k}}{r-1}$
= $\frac{ar^{k}(1+r-1)-a}{r-1}$
= $\frac{ar^{k}(r)-a}{r-1}$
= $\frac{a(r^{k+1}-1)}{r-1}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
13. $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2n+1}{n^{2}} \right )$ $=(n+1)^{2}$
Let P(n) be the statement $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2n+1}{n^{2}} \right )$ $=(n+1)^{2}$
P(1) ⇒ 4=4
P(1) is true
Let P(k) be true
⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $=(k+1)^{2}$
P(k+1) ⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $\left ( 1+\frac{2(k+1)+1}{(k+1)^{2}} \right )$ $=(k+1+1)^{2}$
⇒ $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )...$ $\left ( 1+\frac{2k+1}{k^{2}} \right )$ $\left ( 1+\frac{2k+3}{(k+1)^{2}} \right )$ $=(k+2)^{2}$
L.H.S.,
= $(k+1)^{2}$ $\left ( 1+\frac{2k+3}{(k+1)^{2}} \right )$
= $(k+1)^{2}$ $\frac{(k+1)^{2}+2k+3}{(k+1)^{2}}$
= $k^{2}+4k+4$
= $(k+2)(k+2)$
= $(k+2)^{2}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
14. $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{n} \right )$ $=n+1$
Let P(n) be the statement $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{n} \right )$ $=n+1$
P(1) ⇒ 2=2
P(1) is true
Let P(k) be true
⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $=k+1$
P(k+1) ⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $\left ( 1+\frac{1}{k+1} \right )$ $=k+1+1$
⇒ $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...$ $\left ( 1+\frac{1}{k} \right )$ $\left ( 1+\frac{1}{k+1} \right )$ $=k+2$
L.H.S.,
= $(k+1)$ $\left ( 1+\frac{1}{k+1} \right )$
= $(k+1)$ $\left ( \frac{k+1+1}{k+1} \right )$
= k+2
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
15. $1^{2}+3^{2}+5^{2}+...$ $+(2n-1)^{2}$ $=\frac{n(2n-1)(2n+1)}{3}$
Let P(n) be the statement $1^{2}+3^{2}+5^{2}+...$ $+(2n-1)^{2}$ $=\frac{n(2n-1)(2n+1)}{3}$
P(1) ⇒ 1=1
P(1) is true
Let P(k) be true
⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $=\frac{k(2k-1)(2k+1)}{3}$
P(k+1) ⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $+(2(k+1)-1)^{2}$ $=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$
⇒ $1^{2}+3^{2}+5^{2}+...$ $+(2k-1)^{2}$ $+(2k+1)^{2}$ $=\frac{(k+1)(2k+1)(2k+3)}{3}$
L.H.S.,
= $\frac{k(2k-1)(2k+1)}{3}$ $+(2k+1)^{2}$
= $\frac{k(2k-1)(2k+1)+3(2k+1)^{2}}{3}$
= $\frac{(2k+1)}{3}[k(2k-1)+3(2k+1)]$
= $\frac{(2k+1)}{3}[2k^{2}+5k+3]$
= $\frac{(2k+1)}{3}[(2k+3)(k+1)]$
= $\frac{(k+1)(2k+1)(2k+3)}{3}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
16. $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3n-2)(3n+1)}$ $=\frac{n}{3n+1}$
Let P(n) be the statement $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3n-2)(3n+1)}$ $=\frac{n}{3n+1}$
P(1) ⇒ 1/4=1/4
P(1) is true
Let P(k) be true
⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $=\frac{k}{3k+1}$
P(k+1) ⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$ $=\frac{k+1}{3(k+1)+1}$
⇒ $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$ $+\frac{1}{(3k-2)(3k+1)}$ $+\frac{1}{(3k+1)(3k+4)}$ $=\frac{k+1}{3k+4}$
L.H.S.,
= $\frac{k}{3k+1}$ $+\frac{1}{(3k+1)(3k+4)}$
= $\frac{k(3k+4)+1}{(3k+1)(3k+4)}$
= $\frac{3k^{2}+4k+1}{(3k+1)(3k+4)}$
= $\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
= $\frac{k+1}{3k+4}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
17. $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2n+1)(2n+3)}$ $=\frac{n}{3(2n+3)}$
Let P(n) be the statement $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2n+1)(2n+3)}$ $=\frac{n}{3(2n+3)}$
P(1) ⇒ 1/15=1/15
P(1) is true
Let P(k) be true
⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $=\frac{k}{3(2k+3)}$
P(k+1) ⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$ $=\frac{k+1}{3(2(k+1)+3)}$
⇒ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...$ $+\frac{1}{(2k+1)(2k+3)}$ $+\frac{1}{(2k+3)(2k+5)}$ $=\frac{k+1}{3(2k+5)}$
L.H.S.,
= $\frac{k}{3(2k+3)}$ $+\frac{1}{(2k+3)(2k+5)}$
= $\frac{k(2k+5)+3}{3(2k+3)(2k+5)}$
= $\frac{2k^{2}+5k+3}{3(2k+3)(2k+5)}$
= $\frac{(2k+3)(k+1)}{3(2k+3)(2k+5)}$
= $\frac{k+1}{3(2k+5)}$
L.H.S.=R.H.S.
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
18. $1+2+3+...+n$ $< \frac{1}{8}(2n+1)^{2}$
Let P(n) be the statement $1+2+3+...+n$ $< \frac{1}{8}(2n+1)^{2}$
P(1) ⇒ $1< \frac{9}{8}$
P(1) is true
Let P(k) be true
⇒ $1+2+3+...+k$ $< \frac{1}{8}(2k+1)^{2}$
P(k+1) ⇒ $1+2+3+...+k+(k+1)$ $< \frac{1}{8}(2(k+1)+1)^{2}$
⇒ $1+2+3+...+k+(k+1)$ $< \frac{1}{8}(2k+3)^{2}$
L.H.S.,
$< \frac{1}{8}(2k+1)^{2}+k+1$
$< \frac{1}{8}(4k^{2}+1+4k+8k+8)$
$< \frac{1}{8}(4k^{2}+12k+9)$
$< \frac{1}{8}(2k+3)(2k+3)$
$< \frac{1}{8}(2k+3)^{2}$
L.H.S. $< \frac{1}{8}(2k+3)^{2}$
⇒ P(k+1) is true
Hence by induction P(n) is true for all n∈N
19. n(n+1)(n+5) is a multiple of 3.
Let P(n) be the statement n(n+1)(n+5)
Let P(k) be true
⇒ k(k+1)(k+5)=3q, for some integer q.
P(k+1) ⇒ (k+1) (k+2) (k+5+1)
= (k+1) (k+2) ((k+5)+1)
= (k+1) (k+2) (k+5)+ (k+1) (k+2)
= k (k+1) (k+5)+2 (k+1) (k+5)+ (k+1) (k+2)
= 3q+2 (k+1) (k+5)+ (k+1) (k+2)
= 3q+ 3k2+15k+12
= 3q+3(k+1) (k+4)
= 3[ q+ (k+1) (k+4)], which is multiple of 3.
⇒ P(k+1) is true
20. $10^{2n-1}+1$ is divisible by 11
Let P(n) be the statement $10^{2n-1}+1$
Let P(k) be true
⇒ $10^{2k-1}+1=11q$ , for some integer q.
⇒ $10^{2k-1}=11q-1$
P(k+1) ⇒ $10^{2(k+1)-1}+1$
= $10^{2k+2-1}+1$
= $10^{2k-1}.10^{2}+1$
= $(11q-1).100+1$
= $11q.100-100+1$
= $11q.100-99$
= $11(100q-9)$ , which is divisible by 11.
⇒ P(k+1) is true
21. $x^{2n}-y^{2n}$ is divisible by x+y
Let P(n) be the statement $x^{2n}-y^{2n}$
Let P(k) be true
⇒ $x^{2k}-y^{2k}=(x+y)f(x,y)$
⇒ $x^{2n}=(x+y)f(x,y)+y^{2n}$
P(k+1) ⇒ $x^{2(k+1)}-y^{2(k+1)}$
= $x^{2k}x^{2}-y^{2k}y^{2}$
= $[(x+y)f(x,y)+y^{2k}]x^{2}-y^{2k}y^{2}$
= $(x+y)f(x,y)x^{2}+y^{2k}x^{2}-y^{2k}y^{2}$
= $(x+y)f(x,y)x^{2}+y^{2k}(x^{2}-y^{2})$
= $(x+y)f(x,y)x^{2}+y^{2k}(x-y)(x+y)$
= $(x+y)\left [f(x,y)x^{2}+y^{2k}(x-y) \right ]$ , which is divisible by (x+y)
⇒ P(k+1) is true
22. $3^{2n+2}-8n-9$ is divisible by 8.
Let P(n) be the statement $3^{2n+2}-8n-9$
Let P(k) be true
⇒ $3^{2k+2}-8k-9=8q$
⇒ $3^{2k+2}=8q+8k+9$
P(k+1) ⇒ $3^{2(k+1)+2}-8(k+1)-9$
= $3^{2k+2}.3^{2}-8k-8-9$
= $[8q+8k+9].9-8k-17$
= $8q.9+8k.9+81-8k-17$
= $8q.9+8k.8+64$
= $8[9q+8k+8]$ , which is divisible by 8
⇒ P(k+1) is true
23. $41^{n}-14^{n}$ is multiple of 27
Let P(n) be the statement $41^{n}-14^{n}$
Let P(k) be true
⇒ $41^{k}-14^{k}=27q$
⇒ $41^{k}=27q+14^{k}$
P(k+1) ⇒ $41^{k+1}-14^{k+1}$
= $41^{k}.41-14^{k}.14$
= $[27q+14^{k}].41-14^{k}.14$
= $27q.41+14^{k}.41-14^{k}.14$
= $27q.41+14^{k}.27$
= $27[41q+14^{k}]$ , which is multiple of 27
⇒ P(k+1) is true
24. $(2n+7)< (n+3)^{2}$
Let P(n) be the statement $(2n+7)< (n+3)^{2}$
P(1) ⇒ 9<16
P(1) is true
Let P(k) be true
⇒ $(2k+7)< (k+3)^{2}$
P(k+1) ⇒ $(2(k+1)+7)< ((k+1)+3)^{2}$
⇒ $(2k+7)+2< (k+4)^{2}$
L.H.S.,
$<(k+3)^{2}+2$
$<k^{2}+6k+11$
Now, $k^{2}+6k+11$ $<k^{2}+8k+16$
∴ $(2(k+1)+7)< (k+4)^{2}$
⇒ P(k+1) is true
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