NLt-1-nsin2-2n-sin22-2n-sin2n-2n-

The correct option is D 1/2 1/n{ sin^2π/2n +sin2π/2n+....sin^2xπ/2π} n→∞ μ nbsp;ε nbsp;1/nsin^2 ( x7/2n ) n→∞ x=0 I=∫_0^1sin^2(7/2n ...

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Definite Integral as Limit of Sum

nLt→∞1/nsin2π...

Question

$n L t \to \infty \frac{1}{n} {{sin}^{2} \frac{π}{2 n} + {sin}^{2} \frac{2 π}{2 n} + \dots + {sin}^{2} \frac{n π}{2 n}} =$

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Similar questions

s i n^{2} \frac{π}{2 n + 1}, s i n^{2} \frac{2 π}{2 n + 1}, s i n^{2} \frac{3 π}{2 n + 1}, . . . . ., s i n^{2} \frac{n π}{2 n + 1}

n L t \to \infty {\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \frac{1}{2 n + 3} \dots + \frac{1}{2 n + n}}

2^{n - 1} sin \frac{π}{n} sin \frac{2 π}{n} . . . sin \frac{n - 1}{n} π

n L t \to \infty [\frac{1^{3}}{n^{4} + 1^{4}} + \frac{2^{3}}{n^{4} + 2^{4}} + \dots + \frac{1}{2 n}] =

n L t \to \infty [\frac{1}{1 + n^{3}} + \frac{4}{8 + n^{3}} + \frac{9}{27 + n^{3}} + \dots . + \frac{1}{2 n}] =

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