If n is odd or even,then the sum of n terms o | 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 odd,let $n = 2m + 1$.The sum is $(1 - 2) + (3 - 4) + \do

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

Vedclass · Answer

(D) The given series is $S = 1 - 2 + 3 - 4 + 5 - 6 + \dots + (-1)^{n-1}n$.Case $I$: If $n$ is odd,let $n = 2m + 1$.The sum is $(1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m) + (2m + 1)$.This simplifies to $(-1) + (-1) + \dots + (-1) 	ext{ (m times)} + (2m + 1) = -m + 2m + 1 = m + 1$.Since $n = 2m + 1$,we have $m = \frac{n - 1}{2}$,so the sum is $\frac{n - 1}{2} + 1 = \frac{n + 1}{2}$.Case $II$: If $n$ is even,let $n = 2m$.The sum is $(1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m)$.This is a sum of $m$ terms,each equal to $-1$,so the sum is $-m$.Since $n = 2m$,we have $m = \frac{n}{2}$,so the sum is $-\frac{n}{2}$.Thus,the sum is $\frac{n + 1}{2}$ if $n$ is odd and $-\frac{n}{2}$ if $n$ is even. Option $(d)$ correctly represents this conditional behavior.

If $n$ is odd or even,then the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ will be

Similar Questions

If $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.,$ then $n$ can be:

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

Let $x, y, z$ be positive real numbers such that $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$. Then $x^3 + y^3 + z^3$ is equal to

Five numbers are in $A.P.$,whose sum is $25$ and product is $2520$. If one of these five numbers is $-\frac{1}{2}$,then the greatest number amongst them is

Find the $9^{th}$ term of the $H.P. 6, 4, 3, \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module