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Question Number 87492 by unknown last updated on 04/Apr/20

$$\frac{2+3^2}{1!+2!+3!+4!}+\frac{3+4^2}{2!+3!+4!+5!}+...+\frac{2013+{2014}^2}{2012!+2013!+2014!+2015!}$$

Answered by mind is power last updated on 04/Apr/20

$$\sum _{k⩾2}\frac{k+{(k+1)}^2}{(k-1)!+k!+(k+1)!+(k+2)!}$$$$=\sum _{k⩾2}\frac{k^2+3k+1}{(k-1)!(k+1)(1+k+k(k+2))}$$$$=\sum _{k⩾2}\frac{k^2+3k+1}{(k-1)!(k+1)(k^2+3k+1)}=\sum _{k⩾2}\frac{1}{(k-1)!(k+1)}=\sum _{k⩾2}\frac{k}{(k+1)!}$$$$=\sum _{k⩾2}\frac{k+1-1}{(k+1)!}=\underset{k=2}{\sum ^{2013}}(\frac{1}{k!}-\frac{1}{(k+1)!})=\frac{1}{2}-\frac{1}{2014!}$$$$$$

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