If `Sigma_(r=1)^(2n) sin^(-1) x^(r )=n pi, then Sigma__(r=1)^(2n) x_(r )` is equal to

Options

A. n
B. 2n
C. `(n(n+1))/(2)`
D. none of these

Explanation

To solve the problem, we need to analyze the given equation and find the required summation.

**Step 1: Understand the given equation**
We are given:
\[ \Sigma_{r=1}^{2n} \sin^{-1}(x^r) = n\pi \]
This implies that the sum of the inverse sine of \(x^r\) from \(r=1\) to \(2n\) equals \(n\pi\).
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