For a series S = 1 -2 + 3, -, 4 … n te | vedclass

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Question

$S=1 \cdot 2+3-4+\ldots \ldots \ldots n$ terms

for n to be even, let $n=2 \mathrm{m}$

$S=1 \cdot 2+3 \cdot 4+\ldots \ldots \quad 2 m$ terms

$

Vedclass · Accepted Answer

Both statements are true, and statement $-1$ is the true explanation of statement $-2$

Vedclass · Answer

$S=1 \cdot 2+3-4+\ldots \ldots \ldots n$ terms

for n to be even, let $n=2 \mathrm{m}$

$S=1 \cdot 2+3 \cdot 4+\ldots \ldots \quad 2 m$ terms

$S=(1-2)+(3-4)+(5-6)+\ldots \ldots \ldots m$ terms

$S=(-1)+(-1)+(-1)+\ldots \ldots \ldots m$ terms

$S=-m=\frac{-n}{2}$

for $n$ to be odd let it be $2 \mathrm{m}+1$ so,

$\mathrm{S}=1-2+3-4+\ldots \ldots \ldots(2 \mathrm{m}+1)$ tems

$S=(1-2)+(3-4)+(5-6)+\ldots \ldots \ldots[(2 m-1)-2 m](2 m+1)$

$-m+2 m+1$

$ = (m + 1)\quad \left\langle {\begin{array}{*{20}{c}}
{n = 2m + 1}\
{m = \frac{{n - 1}}{2}}
\end{array}} ightangle $

$\frac{\mathrm{n}-1}{2}+1$

$=\frac{n+1}{2}$

For a series $S = 1 -2 + 3\, -\, 4 … n$ terms,

Statement $-1$ : Sum of series always dependent on the value of $n$ , i.e. whether it is even or odd.

Statement $-2$ : Sum of series is $-\frac {n}{2}$ when value of $n$ is any even integer

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