Q60 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
Assertion (A) : $1+\frac{2.1}{3.2}+\frac{2.5}{3.6} \frac{1}{4}+\frac{2.5 .8}{3.6 .9} \frac{1}{8}+\ldots \infty=\sqrt[3]{4}$
Reason (R): $|x| \lt 1$,
$(1-x)^{-n}=1+n x+\frac{n(n+1)}{1.2} x^2+\frac{n(n+1)(n+2)}{1.2 .3} x^3+\ldots$
The correct answer is
- Correct answerA(A) and (R) are correct, (R) is the correct explanation of (A)
- B(A) and (R) are correct, but (R) is not the correct explanation of (A)
- C(A) is correct but (R) is not correct
- D(A) is not correct but (R) is correct
Solution
$(1-x)^{-n}=1+n x+\frac{n(n+1)}{1.2} x^2+\frac{n(n+1)(n+2)}{1.23} x^3+\ldots$
Put $x=\frac{1}{2}, n=\frac{2}{3}$ $\begin{aligned} & \left(1-\frac{1}{2}\right)^{\frac{-2}{3}}=1+\frac{2}{3} \cdot \frac{1}{2}+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)}{1.2}\left(\frac{1}{2}\right)^2+\ldots . . \\ & \frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right)\left(\frac{1}{2}\right)^3}{1.2 .3}+\ldots \\ & \sqrt[3]{4}=1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2.5}{3.6} \cdot \frac{1}{4}+\frac{2.5 .8}{3.6 .9} \cdot \frac{1}{8}+\ldots .\infty\end{aligned}$
Put $x=\frac{1}{2}, n=\frac{2}{3}$ $\begin{aligned} & \left(1-\frac{1}{2}\right)^{\frac{-2}{3}}=1+\frac{2}{3} \cdot \frac{1}{2}+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)}{1.2}\left(\frac{1}{2}\right)^2+\ldots . . \\ & \frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right)\left(\frac{1}{2}\right)^3}{1.2 .3}+\ldots \\ & \sqrt[3]{4}=1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2.5}{3.6} \cdot \frac{1}{4}+\frac{2.5 .8}{3.6 .9} \cdot \frac{1}{8}+\ldots .\infty\end{aligned}$
Hence, the correct answer is option: 'A'.