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?
Asked
Active
Viewed 6.3k times
6
davidlowryduda
- 91,687
sbr
- 163
-
1Any equation? ${}$ – Will Jagy Oct 02 '12 at 00:32
-
actually, just the series summation. – sbr Oct 02 '12 at 00:33
-
This more recent post (applied to $r=a=n$) answers your question: Deriving Sum of a Geometric Progression – Anne Bauval Jun 13 '23 at 12:10
2 Answers
21
This is called a geometric series.
$$n(1+n+n^2+\cdots n^{n-1})=n\frac{n^n-1}{n-1}$$
Why?
$$S=1+n+n^2+\cdots n^{n-1}$$
$$nS=n+n^2+n^3+\cdots n^{n}$$
$$S(1-n)=1-n^{n}$$
$$S=\frac{1-n^{n}}{1-n}$$
-
1This 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
4
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} $
Mirac Suzgun
- 41