The index of the power of x occurring in the | vedclass

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(D) The $r^{	ext{th}}$ term from the end in the expansion of $(a + b)^n$ is the $(n - r + 2)^{	ext{th}}$ term from the beginning. Here,$n = 12$ and

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(D) The $r^{	ext{th}}$ term from the end in the expansion of $(a + b)^n$ is the $(n - r + 2)^{	ext{th}}$ term from the beginning. Here,$n = 12$ and $r = 5$,so we need the $(12 - 5 + 2) = 9^{	ext{th}}$ term from the beginning. The general term $T_{k+1}$ is given by ${}^{12}C_k (\frac{x^3}{2})^{12-k} (-\frac{2}{x^2})^k$. For the $9^{	ext{th}}$ term,$k = 8$. $T_9 = {}^{12}C_8 (\frac{x^3}{2})^4 (-\frac{2}{x^2})^8$ $T_9 = {}^{12}C_8 \cdot \frac{x^{12}}{2^4} \cdot \frac{2^8}{x^{16}}$ $T_9 = {}^{12}C_8 \cdot 2^4 \cdot x^{12-16}$ $T_9 = {}^{12}C_8 \cdot 16 \cdot x^{-4}$ Thus,the index of the power of $x$ is $-4$.

The index of the power of $x$ occurring in the $5^{\text{th}}$ term from the end in the expansion of $\left(\frac{x^3}{2} - \frac{2}{x^2}\right)^{12}$ is

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