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

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Question

(d) Given series $S = 1 - 2 + 3 - 4 + 5 - 6.........$

Case $I$ : If $n$ is odd, say $2m + 1$

In this case, the number of positive terms

$ = \

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(a) and (c) both

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(d) Given series $S = 1 - 2 + 3 - 4 + 5 - 6.........$

Case $I$ : If $n$ is odd, say $2m + 1$

In this case, the number of positive terms

$ = \frac{1}{2}(n + 1) = \frac{1}{2}(2m + 1 + 1) = (m + 1)$

and the number of negative terms

$ = (2m + 1) - (m + 1) = m$

Then sum $ = [1 + 3 + 5 + .........{\rm{upto}}\,(m + 1)\;{\rm{terms]}}$

$ - [2 + 4 + 6.......{\rm{upto}}\;m\;{\rm{terms}}]$

$ = \frac{1}{2}(m + 1)[2 + (m + 1 - 1)2] - \frac{m}{2}[4 + (m - 1)2]$

$ = (m + 1)(m + 1 - m) = m + 1 = \frac{1}{2}(n + 1)$.

Case $II$ : If $n$ is even

Sum $ = \left( {1 + 3 + 5......{\rm{upto}}\;\frac{n}{2}\,{\rm{terms}}} \right)$

$ - \left( {2 + 4 + 6....{\rm{upto}}\,\frac{n}{{\rm{2}}}{\rm{terms}}} \right)$

$ = \frac{1}{2}.\;\frac{n}{2}\left[ {2 + \left( {\frac{n}{2} - 1} \right)2} \right] - \frac{1}{2}.\frac{n}{2}\left[ {4 + \left( {\frac{n}{2} - 1} \right)2} \right]$

$ = \frac{1}{4}n[n - (n + 2)] = - \frac{n}{2}$.

Trick : Put $n = \;3,\,4$

${S_1} = 2,\;{S_3} = - \,2,$

which the option $(a)$ and $(c)$ give for $n = 3,4$

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

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