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Question Number 74573 by Raxreedoroid last updated on 26/Nov/19

$$\mathrm{Find}(\mathrm{turn}\mathrm{it}\mathrm{into}\mathrm{non}-\mathrm{segma}\mathrm{expression})$$$$1+\underset{k=1}{\sum ^{n-1}}\frac{{(-1)}^k+3}{2}$$

Commented by mathmax by abdo last updated on 26/Nov/19

$$letS=1+\sum _{k=1}^{n-1}\frac{3+{(-1)}^k}{2}=1+\frac{3}{2}\sum _{k=1}^{n-1}\left(1\right)+\frac{1}{2}\sum _{k=1}^{n-1}{(-1)}^k$$$$=1+\frac{3}{2}(n-1)+\frac{1}{2}(\sum _{k=0}^{n-1}{(-1)}^k-1)$$$$=\frac{3n}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{2}\times \frac{1-{(-1)}^n}{2}=\frac{3n}{2}-1+\frac{1}{4}(1-{(-1)}^n)$$$$=\frac{3n}{2}-\frac{3}{4}-\frac{{(-1)}^n}{4}$$

Answered by mind is power last updated on 26/Nov/19

$$1+\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\frac{{(-1)}^{\mathrm{k}}}{2}+\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\frac{3}{2}$$$$=1+\frac{(-1)}{2}.\frac{-1-{(-1)}^{\mathrm{n}-1}}{2}+\frac{3}{2}(\mathrm{n}-1)$$$$=\frac{{(-1)}^{\mathrm{n}-1}+6\mathrm{n}-3}{4}$$

Commented by Smail last updated on 26/Nov/19

$$1-x^n=(1-x)(1+x+x^2+x^3+...+x^{n-1})$$$$\iff \frac{1-x^n}{1-x}=1+x+x^2+...+x^{n-1}$$$$Thus,\underset{k=0}{\sum ^{n-1}}{x}^k=\frac{1-x^n}{1-x}$$

Commented by JDamian last updated on 26/Nov/19

$$\underset{k=1}{\sum ^{n-1}}{x}^{k}isthesumofa\mathit{g}\mathit{e}\mathit{o}\mathit{m}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathit{p}\mathit{r}\mathit{o}\mathit{g}\mathit{e}\mathit{s}\mathit{s}\mathit{i}\mathit{o}\mathit{n}$$

Commented by Raxreedoroid last updated on 26/Nov/19

$$\mathrm{What}\mathrm{the}\mathrm{step}\mathrm{of}\mathrm{expanding}\underset{k=1}{\sum ^{n-1}}\frac{{(-1)}^k}{2}\mathrm{called}?\mathrm{or}\mathrm{how}\mathrm{it}\mathrm{was}\mathrm{done}?$$

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