Evaluate: $ \sum_{i}^{n+1} n !$

Question

$ \sum_{i}^{n+1} n !$

This sum is suppose to equal $(n+1)!$. For some reason I don't get why. Can anyone explain?

If you mean $\sum_{i=1}^{n+1}n!$ then $\sum_{i=1}^{n+1}n!=n!\sum_{i=1}^{n+1}1$ so... — Ixion, May 20 '17 at 05:06

score 3 · Answer 1 · answered May 20 '17 at 05:05

3

Hint: All the terms are the same; $n!$ does not depend on $i$.

answered May 20 '17 at 05:05

Stahl

2

So n! added n+1 times? That's what we are doing? – helios123 May 20 '17 at 05:09
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@helios123 Yes, that's exactly what the sum is. Do you see why it simplifies to $(n+1)!$ now? – Stahl May 20 '17 at 05:11
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(That is, of course, assuming the starting index for $i$ is $1$.) – Stahl May 20 '17 at 05:13

score 0 · Answer 2 · answered May 20 '17 at 05:09

0

Example: $\sum_{1}^{5}2 = 5 \times 2 = 10$; $\sum_{1}^{n}x = n \times x = nx$ for all integers $n \geq 1$ and all real $x$.

Clearly, we have $$ \sum_{1}^{n+1}n! = (n+1)\times n! = (n+1)!. $$

answered May 20 '17 at 05:09

Yes

2 Answers2