Find the coefficient of
1. $\displaystyle \small x^{5}$ in $\displaystyle \small (x+3)^{8}$
We know that,
general term of expansion $\displaystyle \small (x+3)^{8}$ is,
$\displaystyle \small T_{r+1}=^{8}C_{r}x^{8-r}(3)^{r}$ ...(i)
[general term of expansion $\displaystyle \small (a+b)^{n}$ is $\displaystyle \small T_{r+1}=^{n}C_{r}a^{n-r}b^{r}$ ]
We need the term $\displaystyle \small x^{5}$
∴ $\displaystyle \small x^{8-r}=x^{5}$ ⇒ $\displaystyle \small 8-r=5$
⇒ $\displaystyle \small r=3$
Put $\displaystyle \small r=3$ in (i)
⇒ $\displaystyle \small T_{3+1}=^{8}C_{3}x^{8-3}(3)^{3}$
⇒ $\displaystyle \small T_{4}=^{8}C_{3}x^{5}(3)^{3}$
Coefficient of $\displaystyle \small x^{5}$ is $\displaystyle \small ^{8}C_{3}(3)^{3}=1512$
2. $\displaystyle \small a^{5}b^{7}$ in $\displaystyle \small (a-2b)^{12}$
We know that,
general term of expansion $\displaystyle \small (a-2b)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}a^{12-r}(-2b)^{r}$ ...(i)
We need the term $\displaystyle \small a^{5}$
∴ $\displaystyle \small a^{12-r}=a^{5}$ ⇒ $\displaystyle \small 12-r=5$
⇒ $\displaystyle \small r=7$
Put $\displaystyle \small r=7$ in (i)
$\displaystyle \small T_{7+1}=^{12}C_{7}a^{12-7}(-2b)^{7}$
$\displaystyle \small T_{8}=^{12}C_{7}a^{5}(-2)^{7}b^{7}$
Coefficient of $\displaystyle \small a^{5}b^{7}$ is $\displaystyle \small ^{12}C_{7}(-2)^{7}=-101376$
Write the general term in the expansion of
3. $\displaystyle \small (x^{2}-y)^{6}$
General term of expansion $\displaystyle \small (x^{2}-y)^{6}$ is,
$\displaystyle \small T_{r+1}=^{6}C_{r}(x^{2})^{6-r}(-y)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{6}C_{r}x^{12-2r}(y)^{r}$
4. $\displaystyle \small (x^{2}-yx)^{12},x\neq 0$
General term of expansion $\displaystyle \small (x^{2}-yx)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}(x^{2})^{12-r}(-yx)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{12}C_{r}x^{24-2r}(y)^{r}(x)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{12}C_{r}x^{24-r}(y)^{r}$
5. Find the 4th term in the expansion of $\displaystyle \small (x-2y)^{12}$
We know that,
general term of expansion $\displaystyle \small (x-2y)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}x^{12-r}(-2y)^{r}$
We need 4th term, so put $\displaystyle \small r=3$
$\displaystyle \small T_{3+1}=^{12}C_{3}x^{12-3}(-2y)^{3}$
$\displaystyle \small T_{4}=^{12}C_{3}x^{9}(-2)^{3}y^{3}$
$\displaystyle \small T_{4}=-1760x^{9}y^{3}$
6. Find the 13th term in the expansion of $\displaystyle \small \left (9x-\frac{1}{3\sqrt{x}} \right )^{18}, x\neq 0$
We know that,
general term of expansion $\displaystyle \small \left (9x-\frac{1}{3\sqrt{x}} \right )^{18}$ is,
$\displaystyle \small T_{r+1}=^{18}C_{r}(9x)^{18-r}\left (\frac{1}{3\sqrt{x}} \right)^{r}$
We need 13th term, so put $\displaystyle \small r=12$
$\displaystyle \small T_{12+1}=^{18}C_{12}(9x)^{18-12}\left (\frac{1}{3\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9x)^{6}\left (\frac{1}{3\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9)^{6}x^{6}\left (\frac{-1}{3} \right)^{12}\left
(\frac{1}{\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9)^{6}x^{6}\left (\frac{1}{3^{12}} \right)\left (\frac{1}{x^{6}} \right)$
$\displaystyle \small T_{13}=^{18}C_{12}=18564$
Find the middle terms in the expansions of
7. $\displaystyle \small \left (3-\frac{x^{3}}{6} \right )^{7}$
$n=7$ which is an odd number,
∴ middle terms are, $\displaystyle \small \left ( \frac{7+1}{2} \right )$ and $\displaystyle \small \left ( \frac{7+1}{2}+1 \right )$
[middle terms of an odd number are $\displaystyle \small \left ( \frac{n+1}{2} \right )$ and $\displaystyle \small \left ( \frac{n+1}{2}+1 \right )$]
⇒ 4th and 5th terms
general term of expansion $\displaystyle \small \left (3-\frac{x^{3}}{6} \right )^{7}$ is,
$\displaystyle \small T_{r+1}=^{7}C_{r}(3)^{7-r}\left (\frac{-x^{3}}{6} \right)^{r}$ ...(i)
Put $r=3$ in (i)
⇒ $\displaystyle \small T_{3+1}=^{7}C_{3}(3)^{7-3}\left (\frac{-x^{3}}{6} \right)^{3}$
$\displaystyle \small T_{4}=^{7}C_{3}(3)^{4}\left (\frac{-x^{3}}{6} \right)^{3}$
$\displaystyle \small T_{4}=-^{7}C_{3}(3)^{4}\left (\frac{x^{9}}{(6)^{3}} \right)=-\frac{105}{8}x^{9}$
Put $r=4$ in (i)
⇒ $\displaystyle \small T_{4+1}=^{7}C_{4}(3)^{7-4}\left (\frac{-x^{3}}{6} \right)^{4}$
$\displaystyle \small T_{5}=^{7}C_{4}(3)^{3}\left (\frac{-x^{3}}{6} \right)^{4}$
$\displaystyle \small T_{5}=^{7}C_{3}(3)^{3}\left (\frac{x^{12}}{(6)^{4}} \right)=\frac{35}{48}x^{12}$
8. $\displaystyle \small \left (\frac{x}{3}+9y \right )^{10}$
$n=10$ which is an even number,
∴ middle term is, $\displaystyle \small \left ( \frac{10}{2}+1 \right )$
[middle term of an even number is $\displaystyle \small \left ( \frac{n}{2}+1 \right )$]
⇒ 6th term
general term of expansion $\displaystyle \small \left (\frac{x}{3}+9y \right )^{10}$ is,
$\displaystyle \small T_{r+1}=^{10}C_{r}\left (\frac{x}{3} \right)^{10-r}(9y)^{r}$ ...(i)
Put $r=5$ in (i)
⇒ $\displaystyle \small T_{5+1}=^{10}C_{5}\left (\frac{x}{3} \right)^{10-5}(9y)^{5}$
$\displaystyle \small T_{6}=^{10}C_{5}\left (\frac{x}{3} \right)^{5}(9y)^{5}$
$\displaystyle \small T_{6}=^{10}C_{5}\left (\frac{x^{5}}{(3)^{5}} \right)(9)^{5}y^{5}=61236x^{5}y^{5}$
9. In the expansion of $\displaystyle \small (1+a)^{m+n}$, prove that coefficients of $a^{m}$ and $a^{n}$ are equal.
general term of expansion $\displaystyle \small (1+a)^{m+n}$ is,
$\displaystyle \small T_{r+1}=^{m+n}C_{r}a^{r}$
Put $r=m$ and $r=n$
$\displaystyle \small T_{m+1}=^{m+n}C_{m}a^{m}$ ...(i)
and $\displaystyle \small T_{n+1}=^{m+n}C_{n}a^{n}=^{m+n}C_{m+n-n}a^{n}$$\displaystyle \small = ^{m+n}C_{m}a^{n}$...(ii)
[$^{n}C_{r}= ^{n}C_{n-r}$]
From (i) and (ii), coefficients of $a^{m}$ and $a^{n}$ are equal.
10. The coefficients of the $\displaystyle \small (r-1)^{th}$, $\displaystyle \small r^{th}$ and $\displaystyle \small (r+1)^{th}$ terms in the expansion of $\displaystyle \small (x+1)^{n}$ are in the ratio $1:3:5$. Find $n$ and $r$.
We know that,
$\displaystyle \small T_{r-1}=^{n}C_{r-2}x^{n-r+2}$
$\displaystyle \small T_{r}=^{n}C_{r-1}x^{n-r+1}$
$\displaystyle \small T_{r+1}=^{n}C_{r}x^{n-r}$
Given that,
$\displaystyle \small ^{n}C_{r-2}:^{n}C_{r-1}:^{n}C_{r}=1:3:5$
Consider, $\displaystyle \small \frac{^{n}C_{r-2}}{^{n}C_{r-1}}=\frac{1}{3}$
$\displaystyle \frac{\frac{n!}{(r-2)!(n-(r-2))!}}{\frac{n!}{(r-1)!(n-(r-1))!}}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)!(n-r+1)!}{(r-2)!(n-r+2)!}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)(n-r+1)!}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)}{(n-r+2)}=\frac{1}{3}$
$\displaystyle \small 3r-3=n-r+2$
$\displaystyle \small n-4r+5=0$ ...(i)
Now consider, $\displaystyle \small \frac{^{n}C_{r-1}}{^{n}C_{r}}=\frac{3}{5}$
$\displaystyle \frac{\frac{n!}{(r-1)!(n-(r-1))!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}$
$\displaystyle \small \frac{r!(n-r)!}{(r-1)!(n-r+1)!}=\frac{3}{5}$
$\displaystyle \small \frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}=\frac{3}{5}$
$\displaystyle \small \frac{r}{(n-r+1)}=\frac{3}{5}$
$\displaystyle \small 5r=3n-3r+3$
$\displaystyle \small 3n-8r+3=0$ ...(ii)
Solving (i) and (ii),we get $n=7$ and $r=3$
11. Prove that the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$
Consider, $(1+x)^{2n}$
$\displaystyle \small T_{r+1}=^{2n}C_{r}x^{r}$
Put $r=n$,
⇒ $\displaystyle \small T_{n+1}=^{2n}C_{n}x^{n}$
Coefficient of $x^{n}$ is $\displaystyle \small ^{2n}C_{n}$
$\displaystyle \small =\frac{(2n)!}{n!(2n-n)!}$
$\displaystyle \small =\frac{(2n)!}{n!n!}=\frac{(2n)(2n-1)!}{n(n-1)!n!}$
$\displaystyle \small =2\frac{(2n-1)!}{(n-1)!n!}$ ...(i)
Now consider, $(1+x)^{2n-1}$
$\displaystyle \small T_{r+1}=^{2n-1}C_{r}x^{r}$
Put $r=n$,
⇒ $\displaystyle \small T_{n+1}=^{2n-1}C_{n}x^{n}$
Coefficient of $x^{n}$ is $\displaystyle \small ^{2n-1}C_{n}$
$\displaystyle \small =\frac{(2n-1)!}{n!(2n-1-n)!}$
$\displaystyle \small =\frac{(2n-1)!}{n!(n-1)!}$ ...(ii)
From (i) and (ii), it is clear that coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$
12. Find a positive value of $m$ for which the coefficient of $x^{2}$ in the expansion $\displaystyle \small (1+x)^{m}$ is $6$.
general term of expansion $\displaystyle \small (1+x)^{m}$ is,
$\displaystyle \small T_{r+1}=^{m}C_{r}x^{r}$
Put $r=2$,
$\displaystyle \small T_{2+1}=^{m}C_{2}x^{2}$
Given that the coefficient of $x^{2}$ is 6.
⇒ $\displaystyle \small ^{m}C_{2}=6$
$\displaystyle \small \frac{m!}{2!(m-2)!}=6$
$\displaystyle \small \frac{m(m-1)(m-2)!}{2!(m-2)!}=6$
$\displaystyle \small \frac{m(m-1)}{2}=6$
$\displaystyle \small m(m-1)=12$
$\displaystyle \small m^{2}-m=12$
$\displaystyle \small m^{2}-m-12=0$
By solving this quadrstic equation we get, $m=4$ and $m=-3$
coefficient of $x^{2}$ is 6 for a positive value of $m$
∴ $m=4$
1. $\displaystyle \small x^{5}$ in $\displaystyle \small (x+3)^{8}$
We know that,
general term of expansion $\displaystyle \small (x+3)^{8}$ is,
$\displaystyle \small T_{r+1}=^{8}C_{r}x^{8-r}(3)^{r}$ ...(i)
[general term of expansion $\displaystyle \small (a+b)^{n}$ is $\displaystyle \small T_{r+1}=^{n}C_{r}a^{n-r}b^{r}$ ]
We need the term $\displaystyle \small x^{5}$
∴ $\displaystyle \small x^{8-r}=x^{5}$ ⇒ $\displaystyle \small 8-r=5$
⇒ $\displaystyle \small r=3$
Put $\displaystyle \small r=3$ in (i)
⇒ $\displaystyle \small T_{3+1}=^{8}C_{3}x^{8-3}(3)^{3}$
⇒ $\displaystyle \small T_{4}=^{8}C_{3}x^{5}(3)^{3}$
Coefficient of $\displaystyle \small x^{5}$ is $\displaystyle \small ^{8}C_{3}(3)^{3}=1512$
2. $\displaystyle \small a^{5}b^{7}$ in $\displaystyle \small (a-2b)^{12}$
We know that,
general term of expansion $\displaystyle \small (a-2b)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}a^{12-r}(-2b)^{r}$ ...(i)
We need the term $\displaystyle \small a^{5}$
∴ $\displaystyle \small a^{12-r}=a^{5}$ ⇒ $\displaystyle \small 12-r=5$
⇒ $\displaystyle \small r=7$
Put $\displaystyle \small r=7$ in (i)
$\displaystyle \small T_{7+1}=^{12}C_{7}a^{12-7}(-2b)^{7}$
$\displaystyle \small T_{8}=^{12}C_{7}a^{5}(-2)^{7}b^{7}$
Coefficient of $\displaystyle \small a^{5}b^{7}$ is $\displaystyle \small ^{12}C_{7}(-2)^{7}=-101376$
Write the general term in the expansion of
3. $\displaystyle \small (x^{2}-y)^{6}$
General term of expansion $\displaystyle \small (x^{2}-y)^{6}$ is,
$\displaystyle \small T_{r+1}=^{6}C_{r}(x^{2})^{6-r}(-y)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{6}C_{r}x^{12-2r}(y)^{r}$
4. $\displaystyle \small (x^{2}-yx)^{12},x\neq 0$
General term of expansion $\displaystyle \small (x^{2}-yx)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}(x^{2})^{12-r}(-yx)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{12}C_{r}x^{24-2r}(y)^{r}(x)^{r}$
$\displaystyle \small T_{r+1}=((-1)^{r})^{12}C_{r}x^{24-r}(y)^{r}$
5. Find the 4th term in the expansion of $\displaystyle \small (x-2y)^{12}$
We know that,
general term of expansion $\displaystyle \small (x-2y)^{12}$ is,
$\displaystyle \small T_{r+1}=^{12}C_{r}x^{12-r}(-2y)^{r}$
We need 4th term, so put $\displaystyle \small r=3$
$\displaystyle \small T_{3+1}=^{12}C_{3}x^{12-3}(-2y)^{3}$
$\displaystyle \small T_{4}=^{12}C_{3}x^{9}(-2)^{3}y^{3}$
$\displaystyle \small T_{4}=-1760x^{9}y^{3}$
6. Find the 13th term in the expansion of $\displaystyle \small \left (9x-\frac{1}{3\sqrt{x}} \right )^{18}, x\neq 0$
We know that,
general term of expansion $\displaystyle \small \left (9x-\frac{1}{3\sqrt{x}} \right )^{18}$ is,
$\displaystyle \small T_{r+1}=^{18}C_{r}(9x)^{18-r}\left (\frac{1}{3\sqrt{x}} \right)^{r}$
We need 13th term, so put $\displaystyle \small r=12$
$\displaystyle \small T_{12+1}=^{18}C_{12}(9x)^{18-12}\left (\frac{1}{3\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9x)^{6}\left (\frac{1}{3\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9)^{6}x^{6}\left (\frac{-1}{3} \right)^{12}\left
(\frac{1}{\sqrt{x}} \right)^{12}$
$\displaystyle \small T_{13}=^{18}C_{12}(9)^{6}x^{6}\left (\frac{1}{3^{12}} \right)\left (\frac{1}{x^{6}} \right)$
$\displaystyle \small T_{13}=^{18}C_{12}=18564$
Find the middle terms in the expansions of
7. $\displaystyle \small \left (3-\frac{x^{3}}{6} \right )^{7}$
$n=7$ which is an odd number,
∴ middle terms are, $\displaystyle \small \left ( \frac{7+1}{2} \right )$ and $\displaystyle \small \left ( \frac{7+1}{2}+1 \right )$
[middle terms of an odd number are $\displaystyle \small \left ( \frac{n+1}{2} \right )$ and $\displaystyle \small \left ( \frac{n+1}{2}+1 \right )$]
⇒ 4th and 5th terms
general term of expansion $\displaystyle \small \left (3-\frac{x^{3}}{6} \right )^{7}$ is,
$\displaystyle \small T_{r+1}=^{7}C_{r}(3)^{7-r}\left (\frac{-x^{3}}{6} \right)^{r}$ ...(i)
Put $r=3$ in (i)
⇒ $\displaystyle \small T_{3+1}=^{7}C_{3}(3)^{7-3}\left (\frac{-x^{3}}{6} \right)^{3}$
$\displaystyle \small T_{4}=^{7}C_{3}(3)^{4}\left (\frac{-x^{3}}{6} \right)^{3}$
$\displaystyle \small T_{4}=-^{7}C_{3}(3)^{4}\left (\frac{x^{9}}{(6)^{3}} \right)=-\frac{105}{8}x^{9}$
Put $r=4$ in (i)
⇒ $\displaystyle \small T_{4+1}=^{7}C_{4}(3)^{7-4}\left (\frac{-x^{3}}{6} \right)^{4}$
$\displaystyle \small T_{5}=^{7}C_{4}(3)^{3}\left (\frac{-x^{3}}{6} \right)^{4}$
$\displaystyle \small T_{5}=^{7}C_{3}(3)^{3}\left (\frac{x^{12}}{(6)^{4}} \right)=\frac{35}{48}x^{12}$
8. $\displaystyle \small \left (\frac{x}{3}+9y \right )^{10}$
$n=10$ which is an even number,
∴ middle term is, $\displaystyle \small \left ( \frac{10}{2}+1 \right )$
[middle term of an even number is $\displaystyle \small \left ( \frac{n}{2}+1 \right )$]
⇒ 6th term
general term of expansion $\displaystyle \small \left (\frac{x}{3}+9y \right )^{10}$ is,
$\displaystyle \small T_{r+1}=^{10}C_{r}\left (\frac{x}{3} \right)^{10-r}(9y)^{r}$ ...(i)
Put $r=5$ in (i)
⇒ $\displaystyle \small T_{5+1}=^{10}C_{5}\left (\frac{x}{3} \right)^{10-5}(9y)^{5}$
$\displaystyle \small T_{6}=^{10}C_{5}\left (\frac{x}{3} \right)^{5}(9y)^{5}$
$\displaystyle \small T_{6}=^{10}C_{5}\left (\frac{x^{5}}{(3)^{5}} \right)(9)^{5}y^{5}=61236x^{5}y^{5}$
9. In the expansion of $\displaystyle \small (1+a)^{m+n}$, prove that coefficients of $a^{m}$ and $a^{n}$ are equal.
general term of expansion $\displaystyle \small (1+a)^{m+n}$ is,
$\displaystyle \small T_{r+1}=^{m+n}C_{r}a^{r}$
Put $r=m$ and $r=n$
$\displaystyle \small T_{m+1}=^{m+n}C_{m}a^{m}$ ...(i)
and $\displaystyle \small T_{n+1}=^{m+n}C_{n}a^{n}=^{m+n}C_{m+n-n}a^{n}$$\displaystyle \small = ^{m+n}C_{m}a^{n}$...(ii)
[$^{n}C_{r}= ^{n}C_{n-r}$]
From (i) and (ii), coefficients of $a^{m}$ and $a^{n}$ are equal.
10. The coefficients of the $\displaystyle \small (r-1)^{th}$, $\displaystyle \small r^{th}$ and $\displaystyle \small (r+1)^{th}$ terms in the expansion of $\displaystyle \small (x+1)^{n}$ are in the ratio $1:3:5$. Find $n$ and $r$.
We know that,
$\displaystyle \small T_{r-1}=^{n}C_{r-2}x^{n-r+2}$
$\displaystyle \small T_{r}=^{n}C_{r-1}x^{n-r+1}$
$\displaystyle \small T_{r+1}=^{n}C_{r}x^{n-r}$
Given that,
$\displaystyle \small ^{n}C_{r-2}:^{n}C_{r-1}:^{n}C_{r}=1:3:5$
Consider, $\displaystyle \small \frac{^{n}C_{r-2}}{^{n}C_{r-1}}=\frac{1}{3}$
$\displaystyle \frac{\frac{n!}{(r-2)!(n-(r-2))!}}{\frac{n!}{(r-1)!(n-(r-1))!}}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)!(n-r+1)!}{(r-2)!(n-r+2)!}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)(n-r+1)!}=\frac{1}{3}$
$\displaystyle \small \frac{(r-1)}{(n-r+2)}=\frac{1}{3}$
$\displaystyle \small 3r-3=n-r+2$
$\displaystyle \small n-4r+5=0$ ...(i)
Now consider, $\displaystyle \small \frac{^{n}C_{r-1}}{^{n}C_{r}}=\frac{3}{5}$
$\displaystyle \frac{\frac{n!}{(r-1)!(n-(r-1))!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}$
$\displaystyle \small \frac{r!(n-r)!}{(r-1)!(n-r+1)!}=\frac{3}{5}$
$\displaystyle \small \frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}=\frac{3}{5}$
$\displaystyle \small \frac{r}{(n-r+1)}=\frac{3}{5}$
$\displaystyle \small 5r=3n-3r+3$
$\displaystyle \small 3n-8r+3=0$ ...(ii)
Solving (i) and (ii),we get $n=7$ and $r=3$
11. Prove that the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$
Consider, $(1+x)^{2n}$
$\displaystyle \small T_{r+1}=^{2n}C_{r}x^{r}$
Put $r=n$,
⇒ $\displaystyle \small T_{n+1}=^{2n}C_{n}x^{n}$
Coefficient of $x^{n}$ is $\displaystyle \small ^{2n}C_{n}$
$\displaystyle \small =\frac{(2n)!}{n!(2n-n)!}$
$\displaystyle \small =\frac{(2n)!}{n!n!}=\frac{(2n)(2n-1)!}{n(n-1)!n!}$
$\displaystyle \small =2\frac{(2n-1)!}{(n-1)!n!}$ ...(i)
Now consider, $(1+x)^{2n-1}$
$\displaystyle \small T_{r+1}=^{2n-1}C_{r}x^{r}$
Put $r=n$,
⇒ $\displaystyle \small T_{n+1}=^{2n-1}C_{n}x^{n}$
Coefficient of $x^{n}$ is $\displaystyle \small ^{2n-1}C_{n}$
$\displaystyle \small =\frac{(2n-1)!}{n!(2n-1-n)!}$
$\displaystyle \small =\frac{(2n-1)!}{n!(n-1)!}$ ...(ii)
From (i) and (ii), it is clear that coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$
12. Find a positive value of $m$ for which the coefficient of $x^{2}$ in the expansion $\displaystyle \small (1+x)^{m}$ is $6$.
general term of expansion $\displaystyle \small (1+x)^{m}$ is,
$\displaystyle \small T_{r+1}=^{m}C_{r}x^{r}$
Put $r=2$,
$\displaystyle \small T_{2+1}=^{m}C_{2}x^{2}$
Given that the coefficient of $x^{2}$ is 6.
⇒ $\displaystyle \small ^{m}C_{2}=6$
$\displaystyle \small \frac{m!}{2!(m-2)!}=6$
$\displaystyle \small \frac{m(m-1)(m-2)!}{2!(m-2)!}=6$
$\displaystyle \small \frac{m(m-1)}{2}=6$
$\displaystyle \small m(m-1)=12$
$\displaystyle \small m^{2}-m=12$
$\displaystyle \small m^{2}-m-12=0$
By solving this quadrstic equation we get, $m=4$ and $m=-3$
coefficient of $x^{2}$ is 6 for a positive value of $m$
∴ $m=4$
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