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2-3-2-1-2-3-4-3-4-2-2-3-4-5-2013-2014-2-2012-2013-2014-2015-




Question Number 87492 by unknown last updated on 04/Apr/20
((2+3^2 )/(1!+2!+3!+4!))+((3+4^2 )/(2!+3!+4!+5!))+...+((2013+2014^2 )/(2012!+2013!+2014!+2015!))
$$\frac{\mathrm{2}+\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+\frac{\mathrm{3}+\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}+…+\frac{\mathrm{2013}+\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!+\mathrm{2013}!+\mathrm{2014}!+\mathrm{2015}!} \\ $$
Answered by mind is power last updated on 04/Apr/20
Σ_(k≥2) ((k+(k+1)^2 )/((k−1)!+k!+(k+1)!+(k+2)!))  =Σ_(k≥2) ((k^2 +3k+1)/((k−1)!(k+1)(1+k+k(k+2))))  =Σ_(k≥2) ((k^2 +3k+1)/((k−1)!(k+1)(k^2 +3k+1)))=Σ_(k≥2) (1/((k−1)!(k+1)))=Σ_(k≥2) (k/((k+1)!))  =Σ_(k≥2) ((k+1−1)/((k+1)!))=Σ_(k=2) ^(2013) ((1/(k!))−(1/((k+1)!)))=(1/2)−(1/(2014!))
$$\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{k}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\left({k}−\mathrm{1}\right)!+{k}!+\left({k}+\mathrm{1}\right)!+\left({k}+\mathrm{2}\right)!} \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}}{\left({k}−\mathrm{1}\right)!\left({k}+\mathrm{1}\right)\left(\mathrm{1}+{k}+{k}\left({k}+\mathrm{2}\right)\right)} \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}}{\left({k}−\mathrm{1}\right)!\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right)}=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!\left({k}+\mathrm{1}\right)}=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{k}}{\left({k}+\mathrm{1}\right)!} \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{k}+\mathrm{1}−\mathrm{1}}{\left({k}+\mathrm{1}\right)!}=\underset{{k}=\mathrm{2}} {\overset{\mathrm{2013}} {\sum}}\left(\frac{\mathrm{1}}{{k}!}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2014}!} \\ $$$$ \\ $$

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