Q47 / 160 · TS EAMCET 2009 · Mathematics

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

  • A
    ${ }^{12} C_6$
  • Correct answer
    B
    ${ }^{12} C_6+2$
  • C
    ${ }^{12} C_6+4$
  • D
    ${ }^{12} C_6+6$

Solution

$\begin{aligned} & \text { Now, }\left(1+x^2\right)^{12}\left(1+x^{12}+x^{24}+x^{36}\right) \\ & \begin{array}{r}=\left[1+{ }^{12} C_1\left(x^2\right)+{ }^{12} C_2\left(x^2\right)^2+{ }^{12} C_3\left(x^2\right)^3\right. \\ \quad+{ }^{12} C_4\left(x^2\right)^4+{ }^{12} C_5\left(x^2\right)^5+{ }^{12} C_6\left(x^2\right)^6 \\ \left.+\ldots+{ }^{12} C_{12}\left(x^2\right)^{12}\right] \times\left(1+x^{12}+x^{24}+x^{36}\right) \\ \text { Coefficient of } x^{24}={ }^{12} C_6+{ }^{12} C_{12}+1 \\ ={ }^{12} C_6+2\end{array}\end{aligned}$

Hence, the correct answer is option: 'B'.

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