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Question Number 76813 by Rio Michael last updated on 30/Dec/19

$$\underset{\mathrm{k}=1}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}=$$$$\mathrm{A}.\mathrm{\infty }$$$$\mathrm{B}.1$$$$\mathrm{C}.-1$$$$\mathrm{D}.0$$

Commented by JDamian last updated on 30/Dec/19

$$D$$

Commented by Rio Michael last updated on 30/Dec/19

$$\mathrm{why}\mathrm{sir}?$$

Commented by JDamian last updated on 30/Dec/19

$$S=-1,0,-1,0,\cdot \cdot \cdot$$$$S_n=\{\begin{array}{l}-1when\mathit{n}isodd\\ 0when\mathit{n}iseven\end{array}$$$$As2\mathit{n}isalwayseven,then{S}_n=0$$

Commented by Rio Michael last updated on 30/Dec/19

$$\mathrm{sir}\mathrm{k}=1\Rightarrow \mathrm{S}=-1$$$$\mathrm{k}=2\Rightarrow \mathrm{S}=1$$$$\mathrm{S}=-1,1,-1,1...?$$

Commented by JDamian last updated on 30/Dec/19

$$Youareconfused-Idepicted{\mathit{S}}_{n}asthe$$$$\mathit{s}\mathit{u}\mathit{m}ofthesquence{a}_n={(-1)}^{n}upto\mathit{n}.But$$$$youareusing{\mathit{S}}_{n}asthesequence{a}_{n}itself.$$$$Thisproblemisaboutthesumofthe$$$${(-1)}^{n}sequence.$$

Commented by Rio Michael last updated on 30/Dec/19

$$\mathrm{thanks}\mathrm{sir}.$$$$$$

Commented by abdomathmax last updated on 30/Dec/19

$$S_n=\sum _{k=1}^{2n}{(-1)}^k=\sum _{k=0}^{2n}{(-1)}^k-1$$$$=\frac{1-{(-1)}^{2n+1}}{1-(-1)}-1=\frac{2}{2}-1=0$$$$sothecorrectanswerisD$$

Answered by ~blr237~ last updated on 30/Dec/19

$${\mathrm{S}}_{\mathrm{n}}=1+\mathrm{x}+...+{\mathrm{x}}^{\mathrm{n}}=\frac{1-{\mathrm{x}}^{\mathrm{n}+1}}{1-\mathrm{x}}$$$$\underset{\mathrm{k}=1}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}=(-1)+\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}=-1+\frac{1-{(-1)}^{2\mathrm{n}+1}}{1-(-1)}=-1+1=0$$

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