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Question Number 10855 by Joel576 last updated on 27/Feb/17

(3/(1!+2!+3!)) + (4/(2!+3!+4!)) + (5/(3!+4!+5!)) + ... + ((2016)/(2014!+2015!+2016!)) = ?

$$\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}\:+\:\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}\:+\:\frac{\mathrm{5}}{\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}\:+\:...\:+\:\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!}\:=\:? \\ $$

Answered by nume1114 last updated on 28/Feb/17

(3/(1!+2!+3!))+(4/(2!+3!+4!))+...+((2016)/(2014!+2015!+2016!)) =Σ_(n=1) ^(2014) ((n+2)/(n!+(n+1)!+(n+2)!)) =Σ_(n=1) ^(2014) ((n+2)/(n!(1+(n+1)+(n+1)(n+2)))) =Σ_(n=1) ^(2014) ((n+2)/(n!(n^2 +4n+4))) =Σ_(n=1) ^(2014) (1/(n!(n+2))) =Σ_(n=1) ^(2014) ((n+1)/((n+2)!)) =Σ_(n=1) ^(2014) (1/((n+1)!))−(1/((n+2)!)) =((1/(2!))−(1/(3!)))+((1/(3!))−(1/(4!)))+((1/(4!))−(1/(5!))) ...((1/(2014!))−(1/(2015!)))+((1/(2015!))−(1/(2016!))) =(1/2)−(1/(2016!))

$$\:\:\:\:\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}+\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+...+\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{{n}+\mathrm{2}}{{n}!+\left({n}+\mathrm{1}\right)!+\left({n}+\mathrm{2}\right)!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{{n}+\mathrm{2}}{{n}!\left(\mathrm{1}+\left({n}+\mathrm{1}\right)+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{{n}+\mathrm{2}}{{n}!\left({n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{4}\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{\mathrm{1}}{{n}!\left({n}+\mathrm{2}\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}\right) \\ $$$$\:\:\:\:\:\:\:...\left(\frac{\mathrm{1}}{\mathrm{2014}!}−\frac{\mathrm{1}}{\mathrm{2015}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2015}!}−\frac{\mathrm{1}}{\mathrm{2016}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2016}!} \\ $$

Commented by Joel576 last updated on 01/Mar/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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