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

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

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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 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:

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