Question

The coefficient of $t^{24}$ in the expansion $(1+t^2{)}^{12}(1+t^{12}\left)\right(1+t^{24})$ is

• A. ${}^{12}{C}_6+2$
• B. ${}^{12}{C}_5$
• C. ${}^{12}{C}_6$
• D. ${}^{12}{C}_7$

Answer 1

The correct option is A ${}^{12}{C}_6+2$

The given series is: $(1+t^2{)}^{12}(1+t^{12}\left)\right(1+t^{24})$

$=(1+t^{12}+t^{24}+t^{36})(1+t^2{)}^{12}$

To obtain the coeff of $t^{24}$ in the above series, in the first bracket, the first term i.e, $1$ should be multiplied by the coeff. of $t^{24}$ in $(1+t^2{)}^{12}$

the second term i.e., $t^{12}$ should be multiplied by the coeff. of $t^{12}$ in $(1+t^2{)}^{12}$

the third term i.e., $t^{24}$ should be multiplied by the coeff. of $t^0$ in $(1+t^2{)}^{12}$

Now, in the series expansion of $(1+t^2{)}^{12}$, the $r^{th}$ term is given by: ${}_r^{12}C(t^2{)}^{12-r}={}_r^{12}C(t{)}^{24-2r}$

For coeff. of $t^0:24-2r=0\Rightarrow r=12$

$\Rightarrow$ coeff. of $t^0={}_{12}^{12}C=1$

For coeff. of $t^{12}:24-2r=12\Rightarrow r=6$

$\Rightarrow$ coeff. of $t^{12}={}_6^{12}C=924$

For coeff. of $t^{24}:24-2r=24\Rightarrow r=0$

$\Rightarrow$ coeff. of $t^{24}={}_0^{12}C=1$

Thus, total coeff. of $t^{24}$ in $(1+t^{12}+t^{24}+t^{36})(1+t^2{)}^{12}=1+924+1=926$.