If $|x| \lt 1$, then the number of terms in the expansion of $\left[\frac{1}{2}\left(1.2+2.3 x+3.4…

If $|x| \lt 1$, then the number of terms in the expansion of $\left[\frac{1}{2}\left(1.2+2.3 x+3.4 x^2+\ldots . . \infty\right)\right]^{-25}$ is
  1. Infinite
  2. 101
  3. 76
  4. 51

Solution

$\because(1-x)^{-1}=1+x+x^2+x^3+x^4+\ldots \infty$
Differentiate w.r.t. $x$ $(1-x)^{-2}=1+2 x+3 x^2+4 x^3+\ldots \infty$
Differentiate again w.r.t. $x$ $\begin{aligned} & 2(1-x)^{-2}=1 \cdot 2+2 \cdot 3 x+3 \cdot 4 x^2+\ldots \infty \\ & \therefore(1-x)^{-3}=\frac{1}{2}\left(1 \cdot 2+2 \cdot 3 x+3 \cdot 4 x^2+\ldots \infty\right) \end{aligned}$
Now, $\left[\frac{1}{2}\left(1 \cdot 2+2 \cdot 3 x+3 \cdot 4 x^2+\ldots \infty\right)\right]^{-25}$ $=\left[(1-x)^{-3}\right]^{-25}=(1-x)^{75}$ $\therefore$ Number of terms $=75+1=76$.

Asked in: AP EAMCET 2024 (22 May Shift 1)

Practice more Binomial Theorem questions on Aicharya