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) $(e+1)^{2}$

#MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

sorry the correct option is C)

sayan chowdhury - 8 years, 3 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

sayan chowdhury - 8 years, 3 months ago

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

Francisco Rivera - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$