Write the first five terms of each of the sequences whose $\displaystyle n^{th}$ terms are:
1. $\displaystyle a_{n}=n(n+2)$
Given $\displaystyle a_{n}=n(n+2)$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=1(1+2)=3$
$\displaystyle a_{2}=2(2+2)=8$
$\displaystyle a_{3}=3(3+2)=15$
$\displaystyle a_{4}=4(4+2)=24$
$\displaystyle a_{5}=5(5+2)=35$
Therefore, the first five terms are 3,8,15,24,35
2. $\displaystyle a_{n}=\frac{n}{n+1}$
Given $\displaystyle a_{n}=\frac{n}{n+1}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=\frac{1}{1+1}=\frac{1}{2}$
$\displaystyle a_{2}=\frac{2}{2+1}=\frac{2}{3}$
$\displaystyle a_{3}=\frac{3}{3+1}=\frac{3}{4}$
$\displaystyle a_{4}=\frac{4}{4+1}=\frac{4}{5}$
$\displaystyle a_{5}=\frac{5}{5+1}=\frac{5}{6}$
Therefore, the first five terms are $\displaystyle \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$
3. $\displaystyle a_{n}=2^{n}$
Given $\displaystyle a_{n}=2^{n}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=2^{1}=2$
$\displaystyle a_{2}=2^{2}=4$
$\displaystyle a_{3}=2^{3}=8$
$\displaystyle a_{4}=2^{4}=16$
$\displaystyle a_{5}=2^{5}=32$
Therefore, the first five terms are 2,4,8,16,32
4. $\displaystyle a_{n}=\frac{2n-3}{6}$
Given $\displaystyle a_{n}=\frac{2n-3}{6}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=\frac{2*1-3}{6}=\frac{-1}{6}$
$\displaystyle a_{2}=\frac{2*2-3}{6}=\frac{1}{6}$
$\displaystyle a_{3}=\frac{2*3-3}{6}=\frac{1}{2}$
$\displaystyle a_{4}=\frac{2*4-3}{6}=\frac{5}{6}$
$\displaystyle a_{5}=\frac{2*5-3}{6}=\frac{7}{6}$
Therefore, the first five terms are $\displaystyle \frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$
5. $\displaystyle a_{n}=(-1)^{n-1}5^{n+1}$
Given $\displaystyle a_{n}=(-1)^{n-1}5^{n+1}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=(-1)^{1-1}5^{1+1}=5^{2}=25$
$\displaystyle a_{2}=(-1)^{2-1}5^{2+1}=-5^{3}=-125$
$\displaystyle a_{3}=(-1)^{3-1}5^{3+1}=5^{4}=625$
$\displaystyle a_{4}=(-1)^{4-1}5^{4+1}=-5^{5}=-3125$
$\displaystyle a_{5}=(-1)^{5-1}5^{5+1}=5^{6}=15625$
Therefore, the first five terms are 25,-125,625,-3125,15625
6. $\displaystyle a_{n}=n\frac{n^{2}+5}{4}$
Given $\displaystyle a_{n}=n\frac{n^{2}+5}{4}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=1\frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2}$
$\displaystyle a_{2}=2\frac{2^{2}+5}{4}=\frac{18}{4}=\frac{9}{2}$
$\displaystyle a_{3}=3\frac{3^{2}+5}{4}=\frac{42}{4}=\frac{21}{2}$
$\displaystyle a_{4}=4\frac{4^{2}+5}{4}=\frac{84}{4}=21$
$\displaystyle a_{5}=5\frac{5^{2}+5}{4}=\frac{150}{4}=\frac{75}{2}$
Therefore, the first five terms are $\displaystyle \frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$
Find the indicated terms in each of the sequences whose $\displaystyle n^{th}$ terms are:
7. $\displaystyle a_{n}=4n-3;a_{17},a_{24}$
Given $\displaystyle a_{n}=4n-3$
Putting n=17,24 we get,
$\displaystyle a_{17}=4*17-3=65$
$\displaystyle a_{24}=4*24-3=93$
8. $\displaystyle a_{n}=\frac{n^{2}}{2^{n}};a_{7}$
Given $\displaystyle a_{n}=\frac{n^{2}}{2^{n}}$
Putting n=7 we get,
$\displaystyle a_{7}=\frac{7^{2}}{2^{7}}=\frac{49}{128}$
9. $\displaystyle a_{n}=(-1)^{n-1}n^{3};a_{9}$
Given $\displaystyle a_{n}=(-1)^{n-1}n^{3}$
Putting n=9 we get,
$\displaystyle a_{9}=(-1)^{9-1}9^{3}=729$
10. $\displaystyle a_{n}=\frac{n(n-2)}{n+3};a_{20}$
Given $\displaystyle a_{n}=\frac{n(n-2)}{n+3}$
Putting n=20 we get
$\displaystyle a_{20}=\frac{20(20-2)}{20+3}=\frac{360}{23}$
Write the first five terms of each of the sequences and obtain the corresponding series:
11. $\displaystyle a_{1}=3,a_{n}=3a_{n-1}+2$ for all n>1
Given $\displaystyle a_{n}=3a_{n-1}+2$
Putting n=2,3,4,5 we get,
$\displaystyle a_{2}=3a_{2-1}+2=3a_{1}+2$ $\displaystyle =3*3+2=11$
$\displaystyle a_{3}=3a_{3-1}+2=3a_{2}+2$ $\displaystyle =3*11+2=35$
$\displaystyle a_{4}=3a_{4-1}+2=3a_{3}+2$ $\displaystyle =3*35+2=107$
$\displaystyle a_{5}=3a_{5-1}+2=3a_{4}+2$ $\displaystyle =3*107+2=323$
First five terms are 3,11,35,107,323
corresponding series is 3+11+35+107+323+...
12. $\displaystyle a_{1}=-1,a_{n}=\frac{a_{n-1}}{n}, n\geq 2$
Given $\displaystyle a_{n}=\frac{a_{n-1}}{n}$
Putting n=2,3,4,5 we get,
$\displaystyle a_{2}=\frac{a_{2-1}}{2}=\frac{a_{1}}{2}=\frac{-1}{2}$
$\displaystyle a_{3}=\frac{a_{3-1}}{3}=\frac{a_{2}}{3}=\frac{-1/2}{2}=\frac{-1}{6}$
$\displaystyle a_{4}=\frac{a_{4-1}}{4}=\frac{a_{3}}{4}=\frac{-1/6}{4}=\frac{-1}{24}$
$\displaystyle a_{5}=\frac{a_{5-1}}{5}=\frac{a_{4}}{5}=\frac{-1/24}{5}=\frac{-1}{120}$
First five terms are $\displaystyle -1,\frac{-1}{2},\frac{-1}{6},\frac{-1}{24},\frac{-1}{120}$
corresponding series is $\displaystyle -1+\frac{-1}{2}+\frac{-1}{6}+\frac{-1}{24}+\frac{-1}{120}+...$
13. $\displaystyle a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n>2$
Given $\displaystyle a_{n}=a_{n-1}-1$
Putting n=3,4,5 we get,
$\displaystyle a_{3}=a_{3-1}-1=a_{2}-1=2-1=1$
$\displaystyle a_{4}=a_{4-1}-1=a_{3}-1=1-1=0$
$\displaystyle a_{5}=a_{5-1}-1=a_{4}-1=-1$
First five terms are 2,2,1,0,-1
corresponding series is 2+2+1+0+(-1)+...
14. The Fibonacci sequence is defined by $\displaystyle a_{1}=a_{2}=1,a_{n}=a_{n-1}+a_{n-2}$ $\displaystyle ,n>2$ Find $\displaystyle \frac{a_{n+1}}{a_{n}}$ for n=1,2,3,4,5
Given $\displaystyle a_{1}=a_{2}=1,a_{n}=a_{n-1}+a_{n-2}$
Putting n=3,4,5,6 we get,
$\displaystyle a_{3}=a_{3-1}+a_{3-2}=a_{2}+a_{1}$ $\displaystyle =1+1=2$
$\displaystyle a_{4}=a_{4-1}+a_{4-2}=a_{3}+a_{2}$ $\displaystyle =2+1=3$
$\displaystyle a_{5}=a_{5-1}+a_{5-2}=a_{4}+a_{3}$ $\displaystyle =3+2=5$
$\displaystyle a_{6}=a_{6-1}+a_{6-2}=a_{5}+a_{4}$ $\displaystyle =5+3=8$
Now, $\displaystyle \frac{a_{n+1}}{a_{n}}$
for n=1, $\displaystyle \frac{a_{1+1}}{a_{1}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}=1$
for n=2, $\displaystyle \frac{a_{2+1}}{a_{2}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}=2$
for n=3, $\displaystyle \frac{a_{3+1}}{a_{3}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}$
for n=4, $\displaystyle \frac{a_{4+1}}{a_{4}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}$
for n=5, $\displaystyle \frac{a_{5+1}}{a_{5}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}$
1. $\displaystyle a_{n}=n(n+2)$
Given $\displaystyle a_{n}=n(n+2)$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=1(1+2)=3$
$\displaystyle a_{2}=2(2+2)=8$
$\displaystyle a_{3}=3(3+2)=15$
$\displaystyle a_{4}=4(4+2)=24$
$\displaystyle a_{5}=5(5+2)=35$
Therefore, the first five terms are 3,8,15,24,35
2. $\displaystyle a_{n}=\frac{n}{n+1}$
Given $\displaystyle a_{n}=\frac{n}{n+1}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=\frac{1}{1+1}=\frac{1}{2}$
$\displaystyle a_{2}=\frac{2}{2+1}=\frac{2}{3}$
$\displaystyle a_{3}=\frac{3}{3+1}=\frac{3}{4}$
$\displaystyle a_{4}=\frac{4}{4+1}=\frac{4}{5}$
$\displaystyle a_{5}=\frac{5}{5+1}=\frac{5}{6}$
Therefore, the first five terms are $\displaystyle \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$
3. $\displaystyle a_{n}=2^{n}$
Given $\displaystyle a_{n}=2^{n}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=2^{1}=2$
$\displaystyle a_{2}=2^{2}=4$
$\displaystyle a_{3}=2^{3}=8$
$\displaystyle a_{4}=2^{4}=16$
$\displaystyle a_{5}=2^{5}=32$
Therefore, the first five terms are 2,4,8,16,32
4. $\displaystyle a_{n}=\frac{2n-3}{6}$
Given $\displaystyle a_{n}=\frac{2n-3}{6}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=\frac{2*1-3}{6}=\frac{-1}{6}$
$\displaystyle a_{2}=\frac{2*2-3}{6}=\frac{1}{6}$
$\displaystyle a_{3}=\frac{2*3-3}{6}=\frac{1}{2}$
$\displaystyle a_{4}=\frac{2*4-3}{6}=\frac{5}{6}$
$\displaystyle a_{5}=\frac{2*5-3}{6}=\frac{7}{6}$
Therefore, the first five terms are $\displaystyle \frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$
5. $\displaystyle a_{n}=(-1)^{n-1}5^{n+1}$
Given $\displaystyle a_{n}=(-1)^{n-1}5^{n+1}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=(-1)^{1-1}5^{1+1}=5^{2}=25$
$\displaystyle a_{2}=(-1)^{2-1}5^{2+1}=-5^{3}=-125$
$\displaystyle a_{3}=(-1)^{3-1}5^{3+1}=5^{4}=625$
$\displaystyle a_{4}=(-1)^{4-1}5^{4+1}=-5^{5}=-3125$
$\displaystyle a_{5}=(-1)^{5-1}5^{5+1}=5^{6}=15625$
Therefore, the first five terms are 25,-125,625,-3125,15625
6. $\displaystyle a_{n}=n\frac{n^{2}+5}{4}$
Given $\displaystyle a_{n}=n\frac{n^{2}+5}{4}$
Putting n=1,2,3,4,5 we get,
$\displaystyle a_{1}=1\frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2}$
$\displaystyle a_{2}=2\frac{2^{2}+5}{4}=\frac{18}{4}=\frac{9}{2}$
$\displaystyle a_{3}=3\frac{3^{2}+5}{4}=\frac{42}{4}=\frac{21}{2}$
$\displaystyle a_{4}=4\frac{4^{2}+5}{4}=\frac{84}{4}=21$
$\displaystyle a_{5}=5\frac{5^{2}+5}{4}=\frac{150}{4}=\frac{75}{2}$
Therefore, the first five terms are $\displaystyle \frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$
Find the indicated terms in each of the sequences whose $\displaystyle n^{th}$ terms are:
7. $\displaystyle a_{n}=4n-3;a_{17},a_{24}$
Given $\displaystyle a_{n}=4n-3$
Putting n=17,24 we get,
$\displaystyle a_{17}=4*17-3=65$
$\displaystyle a_{24}=4*24-3=93$
8. $\displaystyle a_{n}=\frac{n^{2}}{2^{n}};a_{7}$
Given $\displaystyle a_{n}=\frac{n^{2}}{2^{n}}$
Putting n=7 we get,
$\displaystyle a_{7}=\frac{7^{2}}{2^{7}}=\frac{49}{128}$
9. $\displaystyle a_{n}=(-1)^{n-1}n^{3};a_{9}$
Given $\displaystyle a_{n}=(-1)^{n-1}n^{3}$
Putting n=9 we get,
$\displaystyle a_{9}=(-1)^{9-1}9^{3}=729$
10. $\displaystyle a_{n}=\frac{n(n-2)}{n+3};a_{20}$
Given $\displaystyle a_{n}=\frac{n(n-2)}{n+3}$
Putting n=20 we get
$\displaystyle a_{20}=\frac{20(20-2)}{20+3}=\frac{360}{23}$
Write the first five terms of each of the sequences and obtain the corresponding series:
11. $\displaystyle a_{1}=3,a_{n}=3a_{n-1}+2$ for all n>1
Given $\displaystyle a_{n}=3a_{n-1}+2$
Putting n=2,3,4,5 we get,
$\displaystyle a_{2}=3a_{2-1}+2=3a_{1}+2$ $\displaystyle =3*3+2=11$
$\displaystyle a_{3}=3a_{3-1}+2=3a_{2}+2$ $\displaystyle =3*11+2=35$
$\displaystyle a_{4}=3a_{4-1}+2=3a_{3}+2$ $\displaystyle =3*35+2=107$
$\displaystyle a_{5}=3a_{5-1}+2=3a_{4}+2$ $\displaystyle =3*107+2=323$
First five terms are 3,11,35,107,323
corresponding series is 3+11+35+107+323+...
12. $\displaystyle a_{1}=-1,a_{n}=\frac{a_{n-1}}{n}, n\geq 2$
Given $\displaystyle a_{n}=\frac{a_{n-1}}{n}$
Putting n=2,3,4,5 we get,
$\displaystyle a_{2}=\frac{a_{2-1}}{2}=\frac{a_{1}}{2}=\frac{-1}{2}$
$\displaystyle a_{3}=\frac{a_{3-1}}{3}=\frac{a_{2}}{3}=\frac{-1/2}{2}=\frac{-1}{6}$
$\displaystyle a_{4}=\frac{a_{4-1}}{4}=\frac{a_{3}}{4}=\frac{-1/6}{4}=\frac{-1}{24}$
$\displaystyle a_{5}=\frac{a_{5-1}}{5}=\frac{a_{4}}{5}=\frac{-1/24}{5}=\frac{-1}{120}$
First five terms are $\displaystyle -1,\frac{-1}{2},\frac{-1}{6},\frac{-1}{24},\frac{-1}{120}$
corresponding series is $\displaystyle -1+\frac{-1}{2}+\frac{-1}{6}+\frac{-1}{24}+\frac{-1}{120}+...$
13. $\displaystyle a_{1}=a_{2}=2,a_{n}=a_{n-1}-1,n>2$
Given $\displaystyle a_{n}=a_{n-1}-1$
Putting n=3,4,5 we get,
$\displaystyle a_{3}=a_{3-1}-1=a_{2}-1=2-1=1$
$\displaystyle a_{4}=a_{4-1}-1=a_{3}-1=1-1=0$
$\displaystyle a_{5}=a_{5-1}-1=a_{4}-1=-1$
First five terms are 2,2,1,0,-1
corresponding series is 2+2+1+0+(-1)+...
14. The Fibonacci sequence is defined by $\displaystyle a_{1}=a_{2}=1,a_{n}=a_{n-1}+a_{n-2}$ $\displaystyle ,n>2$ Find $\displaystyle \frac{a_{n+1}}{a_{n}}$ for n=1,2,3,4,5
Given $\displaystyle a_{1}=a_{2}=1,a_{n}=a_{n-1}+a_{n-2}$
Putting n=3,4,5,6 we get,
$\displaystyle a_{3}=a_{3-1}+a_{3-2}=a_{2}+a_{1}$ $\displaystyle =1+1=2$
$\displaystyle a_{4}=a_{4-1}+a_{4-2}=a_{3}+a_{2}$ $\displaystyle =2+1=3$
$\displaystyle a_{5}=a_{5-1}+a_{5-2}=a_{4}+a_{3}$ $\displaystyle =3+2=5$
$\displaystyle a_{6}=a_{6-1}+a_{6-2}=a_{5}+a_{4}$ $\displaystyle =5+3=8$
Now, $\displaystyle \frac{a_{n+1}}{a_{n}}$
for n=1, $\displaystyle \frac{a_{1+1}}{a_{1}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}=1$
for n=2, $\displaystyle \frac{a_{2+1}}{a_{2}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}=2$
for n=3, $\displaystyle \frac{a_{3+1}}{a_{3}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}$
for n=4, $\displaystyle \frac{a_{4+1}}{a_{4}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}$
for n=5, $\displaystyle \frac{a_{5+1}}{a_{5}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}$
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