Question

The coefficient of ${\mathrm{t}}^{24}$ in the expansion of ${(1+{\mathrm{t}}^2)}^{12}(1+{\mathrm{t}}^{12})(1+{\mathrm{t}}^{24})$ 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 explanation for the correct option:

Finding the coefficient of ${\mathrm{t}}^{24}$:

Given, that the expression is ${(1+{\mathrm{t}}^2)}^{12}(1+{\mathrm{t}}^{12})(1+{\mathrm{t}}^{24})$

We know that the binomial expansion is ${\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{n}}{\mathbf{b}}^{\mathbf{0}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{1}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{b}}^{\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{2}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}{\mathbf{a}}^{\mathbf{0}}{\mathbf{b}}^{\mathbf{n}}$

Re-writing the given expression using Binomial expression, we get

${(1+t^2)}^{12}(1+t^{12})(1+t^{24})=({}^{12}C_0+{}^{12}C_1{t}^2+{}^{12}C_2{t}^4+{}^{12}C_3{t}^6+\dots +{}^{12}C_6{t}^{12}+\dots +{}^{12}C_{12}{t}^{24})(1+t^{12})(1+t^{24})$

[Since, by using the Binomial expression in ${(1+{\mathrm{t}}^2)}^{12}=({}^{12}\mathrm{C}_0+{}^{12}\mathrm{C}_1{\mathrm{t}}^2+{}^{12}\mathrm{C}_2{\mathrm{t}}^4+{}^{12}\mathrm{C}_3{\mathrm{t}}^6+\dots +{}^{12}\mathrm{C}_6{\mathrm{t}}^{12}+\dots +{}^{12}\mathrm{C}_{12}{\mathrm{t}}^{24})$]

Therefore, the coefficient of ${\mathrm{t}}^{24}$in the given expression can be taken as:

$$\begin{gathered} \Rightarrow {}^{12}\mathrm{C}_6+{}^{12}\mathrm{C}_{12}\times 1\times 1+{}^{12}\mathrm{C}_{0}1\times 1 \\ \Rightarrow {}^{12}\mathrm{C}_6+1+1 \\ \mathrm{Therefore},{}^{12}\mathrm{C}_6+2. \end{gathered}$$

Hence, option(A) is the correct answer.