The value of n -limk - 1n-n-k-n2cos4k-n - 1-a 1 - cos c- then

S = 1/nk = 1n∑ (1 k/n) cos 4(k/n) = ∫^1_0 (1 x) cos 4x dx = [(1 x) sin 4x/4]^1_0 + 1/4∫^1_0sin 4x dx = ( 1/16cos 4x)^1_0 = 1/16(1 cos 4) ...

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Byju's Answer

Standard XIII

Mathematics

Summation Using Sigma

The value of ...

Question

The value of $l i m n \to \infty n \sum k = 1 (\frac{n - k}{n^{2}}) cos \frac{4 k}{n} = \frac{1}{a} (1 - cos c),$ then

a + c = 20

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\frac{a}{c} = 4

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a - c = 12

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a c = 16

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Solution

The correct option is C $a - c = 12$
$S = \frac{1}{n} n \sum k = 1 (1 - \frac{k}{n}) cos 4 (\frac{k}{n}) = \int_{0}^{1} (1 - x) cos 4 x d x$
$= [(1 - x) \frac{sin 4 x}{4}]_{0}^{1} + \frac{1}{4} \int_{0}^{1} sin 4 x d x$
$= (\frac{- 1}{16} cos 4 x)_{0}^{1} = \frac{1}{16} (1 - cos 4)$

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The value of limn→∞n∑k=1(n−kn2)cos4kn=1a(1−cos c), then

The correct option is C a−c=12S=1nn∑k=1(1−kn)cos4(kn)=∫10(1−x)cos4x dx =[(1−x)sin4x4]10+14∫10sin4x dx =(−116cos4x)10=116(1−cos4)

The value of $l i m n \to \infty n \sum k = 1 (\frac{n - k}{n^{2}}) cos \frac{4 k}{n} = \frac{1}{a} (1 - cos c),$ then

The correct option is C $a - c = 12$
$S = \frac{1}{n} n \sum k = 1 (1 - \frac{k}{n}) cos 4 (\frac{k}{n}) = \int_{0}^{1} (1 - x) cos 4 x d x$
$= [(1 - x) \frac{sin 4 x}{4}]_{0}^{1} + \frac{1}{4} \int_{0}^{1} sin 4 x d x$
$= (\frac{- 1}{16} cos 4 x)_{0}^{1} = \frac{1}{16} (1 - cos 4)$