Question

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

• A. ${}^{12}{C}_6+3$
• B. ${}^{12}{C}_6+1$
• C. ${}^{12}{C}_6$
• D. ${}^{12}{C}_6+2$

Answer 1

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

$(1+t^2{)}^{12}(1+t^{12}\left)\right(1+t^{24})=(1+t^{12}+t^{24}+t^{36}\left)\right(1+1^2{)}^{12}\therefore$ Coeff. of $t^{24}=1\times$ Coeff. of $t^{24}$in $(1+t^2{)}^{12}+1\times$Coeff. of
$t^{12}$in $(1+t^2{)}^{12}+1\times$ constant term in $(1+t^2{)}^{12}{=}^{12}{C}_{12}{+}^{12}{C}_6{+}^{12}{C}_0=1{+}^{12}{C}_6+1{=}^{12}{C}_6+2$