$\begingroup$

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?

$\endgroup$ 2

2 Answers

$\begingroup$

This is called a geometric series.

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

Why?

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

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

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

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

$\endgroup$ 1 $\begingroup$

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} $

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy