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) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{a}{c} = 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a - c = 12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a c = 16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. 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

Q. If

S_{1} = \sum_{k = 0}^{n} {^{n} C}_{k} cos (k x) cos (n - k) x

and

S_{2} = \sum_{k = 0}^{n} {^{n} C}_{k} sin (k x) sin (n - k) x

, then

S_{1} - S_{2}

equals

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

If $a_{0}, a_{1}, a_{2}, . . . . . . . . . . . ., a_{n - 1}$ are nth roots of unity, then $a_{k}$ = cos $\frac{2 k π}{n}$ + i sin $\frac{2 k π}{n}$ where $0 \leq k \leq n - 1$ .

Also then Value of $2^{n - 1}$ sin $(\frac{π}{n})$ sin $(\frac{2 π}{n})$ .........sin $(\frac{n - 1}{n} π)$ is

Q. The general value of the real angle

θ,

which satisfies the equation,

(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) . . . (cos n θ + i sin n θ) = 1

is given by, (assuming

k

is an integer)

Join BYJU'S Learning Program

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