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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?

  • 0
    actually, just the series summation.2012-10-02

2 Answers 2

19

This is called a geometric series.

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

Why?

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

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

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

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

  • 0
    This is the case $n \ne 1$, of course. We can also do that case, but by other methods.2015-02-21
2

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} $