Summarization in terms of e

(Brazilian Olympics of Math - Round 1 - High School Level)

Being $e = \sum_{n=0}^\infty 1/n!$ The value in terms of e of $\sum_{n=0}^\infty (n+1)^{2}/n!$

A) $2e$B) $4e$C) $5e$D) $e$E)$(e+1)^{2}$

Note by Victor Colombo
5 years, 4 months ago

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sorry the correct option is C)

- 5 years, 4 months ago

the answer is D) 5e we know :::::(n+1)^2/n! =(n^2+2n+1)/n! =(n^2)/n!+2n/n!+1/n! =n/(n-1)!+2/(n-1)!+1/n! =(n-1+1)/(n-1)!+2/(n-1)!+1/n! =(n-1)/(n-1)!+1/(n-1)!+2/(n-1)!+1/n! =1/(n-2)!+3/(n-1)!+1/n! now you know \sum{i=0}^infinity 1/n!=\sum{i=1}^infinity 1/(n-1)!=\sum_{i=2}^infinity 1/(n-2)!=e
so the summation is 5e

- 5 years, 4 months ago

The answer is indeed $$5e$$, however I believe this option corresponds to $\boxed{C) \, 5e}$

- 5 years, 4 months ago