Question

Coefficient of $t^{24}$ in ${(1+t^2)}^{12}(1+t^{12})(1+t^{24})$ 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

${}^{12}C_6+2$

Explanation for the correct option:

Step 1: Expand the given expression:

As the required coefficient is of $t^{24}$.

Expand the expression to get a simplified form of it.

$\begin{array}{rcl}{(1+t^2)}^{12}(1+t^{12})(1+t^{24})& =& {(1+t^2)}^{12}(1+t^{12}+t^{24}+t^{36})\\ & =& {(1+t^2)}^{12}+{t^{12}(1+t^2)}^{12}+{t^{24}(1+t^2)}^{12}+{t^{36}(1+t^2)}^{12}\end{array}$

Step 2: Use the binomial property to get the coefficient:

Eliminate $\begin{array}{rcl}& & {t^{36}(1+t^2)}^{12}\end{array}$ in finding the coefficient of $t^{24}$.

$\begin{array}{rcl}& & {(1+t^2)}^{12}+{t^{12}(1+t^2)}^{12}+{t^{24}(1+t^2)}^{12}\end{array}$

Estimate coefficient using the binomial distribution.

$\begin{array}{rcl}c& =& {}^{12}C_{12}+{}^{12}C_6+{}^{12}C_0\\ & =& 1+{}^{12}C_6+1\\ & =& {}^{12}C_6+2\end{array}$

Hence, option (D) is the correct answer.