Question-159479

Question Number 159479 by quvonnn last updated on 17/Nov/21

Answered by Tokugami last updated on 17/Nov/21

(2/(4×2!))+(2/(5×3!))+(2/(6×4!))+(2/(7×5!))+(2/(8×6!))+... =Σ_(n=2) ^∞ (2/((n+2)n!)) =2Σ_(n=2) ^∞ ((1(n+1))/((n+2)(n+1)n!)) =2Σ_(n=2) ^∞ ((n+1)/((n+2)!)) n+2→n =2Σ_(n=4) ^∞ ((n−1)/(n!)) =2Σ_(n=4) ^∞ ((n/(n!))−(1/(n!))) =2Σ_(n=4) ^∞ (1/((n−1)!))−2Σ_(n=4) ^∞ (1/(n!)) Σ_(n=4) ^∞ (1/((n−1)!)) n−1→n =2(Σ_(n=3) ^∞ (1/(n!))−Σ_(n=4) ^∞ (1/(n!))) =2((Σ_(n=0) ^∞ (1/(n!))−Σ_(n=0) ^2 (1/(n!)))−(Σ_(n=0) ^∞ (1/(n!))−Σ_(n=0) ^3 (1/(n!)))) =2(−((1/(0!))+(1/(1!))+(1/(2!)))+((1/(0!))+(1/(1!))+(1/(2!))+(1/(3!)))) =(2/(3!))=(2/6) =(1/3)

$$\frac{\mathrm{2}}{\mathrm{4}×\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{5}×\mathrm{3}!}+\frac{\mathrm{2}}{\mathrm{6}×\mathrm{4}!}+\frac{\mathrm{2}}{\mathrm{7}×\mathrm{5}!}+\frac{\mathrm{2}}{\mathrm{8}×\mathrm{6}!}+… \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right){n}!} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$${n}+\mathrm{2}\rightarrow{n} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\:\frac{{n}−\mathrm{1}}{{n}!} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\left(\frac{{n}}{{n}!}−\frac{\mathrm{1}}{{n}!}\right) \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\mathrm{2}\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!} \\ $$$$\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!} \\ $$$${n}−\mathrm{1}\rightarrow{n} \\ $$$$=\mathrm{2}\left(\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}−\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}\right) \\ $$$$=\mathrm{2}\left(\left(\cancel{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}}−\underset{{n}=\mathrm{0}} {\overset{\mathrm{2}} {\sum}}\:\frac{\mathrm{1}}{{n}!}\right)−\left(\cancel{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}}−\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{n}!}\right)\right) \\ $$$$=\mathrm{2}\left(−\left(\cancel{\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}}+\frac{\mathrm{1}}{\mathrm{3}!}\right)\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}!}=\frac{\mathrm{2}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by quvonnn last updated on 17/Nov/21

$${thanh}\:{you} \\ $$

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