Question

Coefficient of $t^{24}in(1+t^2{)}^{12}(1+t^{12}\left)\right(1+t^{24})$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$(1+t^{12})(1+t^{24})=1+t^{36}+t^{12}+t^{24}$
$(1+t^2{)}^{12}(1+t^{12}\left)\right(1+t^{24})=(1+t^{36}+t^{12}+t^{24}\left)\right(1+t^2{)}^{12}$, To find the coefficient of $t^{24}$ in $(1+t^2{)}^{12},t^{36}+t^{12}+t^{24})(1+t^2{)}^{12}$, we will find the coefficient of $t^{24}$ in $(1+t^2{)}^{12},t^{36}(1+t^2{)}^{12},t^{12}(1+t^2{)}^{12}$ and $t^{24}\left)\right(1+t^2{)}^{12}$ and add them.

1) Coefficient of $t^{24}$ in $(1+t^2{)}^{12}$
$T_{r+1}={}^{12}{C}_r(t^2{)}^r$
$2r=24orr=12$
$\Rightarrow$ Coefficient of $t^{24}is{}^{12}{C}_{12}=1$

2)Coefficient of $t^{24}in{t}^{36}(1+t^2{)}^{12}$
Since ther is already $t^{36}$ every term in the expansion will have power of x greater than 24. So the coefficient of $t^{24}$ will be zero.

3) Coefficient of $t^{24}$ in $t^{12}(1+t^2{)}^{12}$
Since there is already $t^{12}$ common, to find the coefficient of $t^{24}$ in $t^{12}(1+t^2{)}^{12}$, we have to find the coefficient of $t^{12}$ in $(1+t^2{)}^{12}$
$T_{r+1}={}^{12}{C}_r(t^2{)}^r$
2r=12 or r=6
$\Rightarrow$ Coefficient of $t^{24}$ is ${}^{12}{C}_6$

4) Coefficient of $t^{24}$ in $t^{24}(1+t^2{)}^{12}$
Here, we have to find the term independent of x because $t^{24}$ is there as a common term. The term independent of x is the first term, which is 1
So the coefficient of $t^{24}$ in $(1+t^{36}+t^{12}+t^{24})(1+t^2{)}^{12}={}^{12}{C}_6+0+1+1$
$={}^{12}{C}_6+2$