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Question Number 116056 by Study last updated on 30/Sep/20

Σ_(n=1) ^∞ (5^n /(n!))=?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{{n}} }{{n}!}=? \\ $$

Answered by Dwaipayan Shikari last updated on 30/Sep/20

Σ_(n=1) ^∞ (x^n /(n!))=e^x −1 Σ_(n=1) ^∞ (5^n /(n!))=e^5 −1

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{e}^{\mathrm{x}} −\mathrm{1} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{e}^{\mathrm{5}} −\mathrm{1} \\ $$

Answered by MWSuSon last updated on 30/Sep/20

Σ_(n=1) ^∞ (5^n /(n!))=((5/1)+(5^2 /(2!))+(5^3 /(3!))+(5^4 /(4!))+...) e^x =1+((x/1)+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!))+...) they are thesame when x=5 Σ_(n=1) ^∞ (5^n /(n!))=e^5 −1

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{n}!}=\left(\frac{\mathrm{5}}{\mathrm{1}}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{5}^{\mathrm{4}} }{\mathrm{4}!}+...\right) \\ $$$$\mathrm{e}^{\mathrm{x}} =\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+...\right) \\ $$$$\mathrm{they}\:\mathrm{are}\:\mathrm{thesame}\:\mathrm{when}\:\mathrm{x}=\mathrm{5} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{e}^{\mathrm{5}} −\mathrm{1} \\ $$$$ \\ $$

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