If n is odd or even,the sum of n terms of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given series is $S = 1 - 2 + 3 - 4 + 5 - 6 + \dots + (-1)^{n-1}n$.Case $I$: If $n$ is even,let $n = 2k$.The sum is $(1-2) + (3-4) + \dots + ((

Vedclass · Accepted Answer

Both $(a)$ and $(c)$ depending on $n$

Vedclass · Answer

(D) The given series is $S = 1 - 2 + 3 - 4 + 5 - 6 + \dots + (-1)^{n-1}n$.Case $I$: If $n$ is even,let $n = 2k$.The sum is $(1-2) + (3-4) + \dots + ((2k-1) - 2k) = (-1) + (-1) + \dots + (-1)$ ($k$ times).Sum $= -k = -\frac{n}{2}$.Case $II$: If $n$ is odd,let $n = 2k+1$.The sum is $(1-2) + (3-4) + \dots + ((2k-1) - 2k) + (2k+1) = -k + (2k+1) = k+1$.Since $n = 2k+1$,we have $k = \frac{n-1}{2}$,so the sum is $\frac{n-1}{2} + 1 = \frac{n+1}{2}$.Thus,the sum is $-\frac{n}{2}$ if $n$ is even and $\frac{n+1}{2}$ if $n$ is odd.

If $n$ is odd or even,the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ is given by:

Similar Questions

If $a, x$ are real numbers and $|a| < 1, |x| < 1$, then $1 + (1+a)x + (1+a+a^2)x^2 + \dots \infty$ is equal to

The sum of $n$ terms of the series $12 + 16 + 24 + 40 + \dots$ is

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

$1 + (1 + a)x + (1 + a + a^2)x^2 + \dots \infty = \dots \, (0 < a, x < 1)$

The sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module