Question:
\(\lim\limits_{n \rightarrow \infin}\frac{6}{n+2}\left\{(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+...+(2+\frac{n-1}{2})^2\right\}\) equals
\(\lim\limits_{n \rightarrow \infin}\frac{6}{n+2}\left\{(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+...+(2+\frac{n-1}{2})^2\right\}\) equals
Updated On: Oct 1, 2024
- 38
- 36
- 32
- 30
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The Correct Option is A
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The correct option is (A) : 38.
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