Summation rules for approaching a constant raises to the power of i?

Question

i have to solve from the left hand side to the right,

$ \sum_{i=1}^n i5^i = \frac{5(4n5^n-5^n+1)}{16}$

my thought is to just convert both i's to

$\frac {n(n+1)}{2}$

then keep solving, but Im wondering if there are any summation tricks or formulas when i is in the exponent position?

Answer 1

Let for $x\ne1$ $$S=\sum_{i=1}^nix^i=x+2x^2+3x^3+\cdots+(n-2)x^{n-2}+(n-1)x^{n-1}+nx^n$$

$$xS=x^2+2x^3+3x^4+\cdots+(n-2)x^{n-1}+(n-1)x^n+nx^{n+1}$$

$$(1-x)S$$ $$=x+x^2(2-1)+x^3(3-2)+\cdots+(n-1-(n-2))x^{n-1}+(n-(n-1))x^n-nx^{n+1}$$

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

$$=-nx^{n+1}+\dfrac{x(1-x^n)}{1-x}$$

Answer 2

It's not that simple...

First, note that:

$\begin{align*} \sum_{1 \le k \le n} z^k &= \frac{z (1 - z^n)}{1 - z} \\ \sum_{1 \le k \le n} k z^k &= z \frac{d}{d z} \frac{z (1 - z^n)}{1 - z} \\ &= \frac{z (1 - (n + 1) z^n + n z^{n + 1})}{(1 - z)^2} \end{align*}$

Everything in sight is a polynomial or a rational function, no troubles there. Replacing $z = 5$ gives:

$\begin{equation*} \sum_{1 \le k \le n} k \cdot 5^k = \dfrac{(20 n - 5) \cdot 5^n + 5}{16} \end{equation*}$

The help of maxima with the algebra is gratefully acknowledged.