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Question:
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Assertion (A) If $|x| < 1$, then
$$
\sum_{n=0}^{\infty}(-1)^n x^{n+1}=\frac{x}{x+1}
$$
Reason (R) If $|x| < 1$, then $(1+x)^{-1}$
$$
=1-x+x^2-x^3+\ldots
$$
Which one of the following is true?
Options:
$$
\sum_{n=0}^{\infty}(-1)^n x^{n+1}=\frac{x}{x+1}
$$
Reason (R) If $|x| < 1$, then $(1+x)^{-1}$
$$
=1-x+x^2-x^3+\ldots
$$
Which one of the following is true?
Solution:
1575 Upvotes
Verified Answer
The correct answer is:
$(A)$ and $(R)$ are true, $(R)$ is a correct explanation of $(A)$
We have,
$$
\begin{aligned}
\frac{x}{x+1} & =x(1+x)^{-1}=x\left(1-x+x^2-x^3+x^4-\ldots\right) \\
& =x-x^2+x^3-x^4+x^5-\ldots \\
& =\sum_{n=0}^{\infty}(-1)^n x^{n+1}
\end{aligned}
$$
$\therefore$ Assertion is true.
Reason $(R)=(1+x)^{-1}=1-x+x^2-x^3+x^4-\ldots$ is also true (A) and (R) are true, (R) is correct explanation of (A).
$$
\begin{aligned}
\frac{x}{x+1} & =x(1+x)^{-1}=x\left(1-x+x^2-x^3+x^4-\ldots\right) \\
& =x-x^2+x^3-x^4+x^5-\ldots \\
& =\sum_{n=0}^{\infty}(-1)^n x^{n+1}
\end{aligned}
$$
$\therefore$ Assertion is true.
Reason $(R)=(1+x)^{-1}=1-x+x^2-x^3+x^4-\ldots$ is also true (A) and (R) are true, (R) is correct explanation of (A).
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