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

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(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

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

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(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

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