# Solving a series $n(1 + n + n^2 + n^3 + n^4 +…n^{n-1})$

I'm trying to sum the following series?

$n(1 + n + n^2 + n^3 + n^4 +.......n^{n-1})$

Do you have any ideas?

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Any equation? ${}$ – Will Jagy Oct 2 '12 at 0:32
actually, just the series summation. – sbr Oct 2 '12 at 0:33

This is called a geometric series.

$$n(1+n+n^2+\cdots n^{n-1})=n\frac{n^n-1}{n-1}$$

Why?

$$S=1+n+n^2+\cdots n^{n-1}$$

$$nS=n+n^2+n^3+\cdots n^{n}$$

$$S(1-n)=1-n^{n}$$

$$S=\frac{1-n^{n}}{1-n}$$

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thanks. awesome. – sbr Oct 2 '12 at 0:39
This is the case $n \ne 1$, of course. We can also do that case, but by other methods. – GEdgar Feb 21 '15 at 16:29

Let $n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n = Sum$

Then,

$1 + n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n = Sum + 1$

$n \times (1 + n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n ) = n \times (Sum + 1)$

$n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n + n^{n+1} = n\times Sum + n$

$(n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n) + n^{n+1} = n\times Sum + n$

$(Sum)+ n^{n+1} = n\times Sum + n$

$n^{n+1} = (n-1) \times Sum + n$

$n^{n+1} -n = (n-1) \times Sum$

$\frac {n^{n+1} -n}{n-1} = Sum$

Hence,

$Sum = \sum_{i = 1}^{n} n^i = \frac {n^{n+1} -n}{n-1}$

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