How would one begin to find a simple formula for
$\displaystyle S_m = \sum_{n=1}^{m} n(2^n)$
I have no idea where to begin
If I recall correctly, the following is a property of sums
$\displaystyle \begin{align*} \sum_{n=1}^{m}n(2^n) &= \sum_{n=1}^{m}n\sum_{n=1}^{m}2^n \end{align*}$
On the RHS, the first sum obviously becomes $\displaystyle \frac{m(m+1)}{2}$
I found the 2nd sum by letting
$\displaystyle \begin{align*}\ S_n &= 2 + 2^2 + 2^3 + ... + 2^n \\ 2S_n &= 2^2 + 2^3 + 2^4 + ... + 2^n + 2^{n+1}\end{align*} $
Subtracting $\displaystyle S_n$ from $\displaystyle 2S_n$ yields
$\displaystyle S_n = 2^{n+1} - 2 = 2(2^n) - 2 = 2(2^n -1)$
Plugging back in...
$\displaystyle \begin{align*} \sum_{n=1}^{m}n\sum_{n=1}^{m}2^n &= \frac{2m(m+1)(2^m - 1)}{2} \\ &= (m^2+m)(2^m - 1)\end{align*}$
Is any of my reasoning is even correct? I have no idea how Wolfram Alpha got $\displaystyle 2(2^{m} m - 2^m + 1)$
(Rofl)Hello, ReneG!
Find a closed-form expression for: .$\displaystyle S_n \:=\: \sum_{i=1}^{n} n(2^n)$
$\displaystyle \begin{array}{ccccccc}\text{We are given:} & S_n &=& 1\cdot2 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot2^4 + \cdots + n\cdot2^n \qquad\qquad\qquad & [1] \\ \text{Mult. by 2:} & 2S_n &=& \qquad\quad 1\cdot2^2 + 2\cdot 2^3 + 3\cdot 2^4 + \cdots + (n-1)2^n + n\cdot2^{n+1} & [2] \\ \text{Subtract [1] - [2]:} & -S_n &=& 2 + 2^2 + 2^3 + 2^4 + \cdots + 2n - n\cdot 2^{n+1}\qquad\qquad\qquad\qquad \end{array}$
$\displaystyle \text{We have: }\: -S_n \;=\; \underbrace{2 + 2^2 + 2^3 + 2^4 + \cdots + 2n}_{\text{geometric series}} - n\cdot 2^{n+1}\;\;[3]$
The geometric series has: first term $\displaystyle a = 2$, common ratio $\displaystyle r = 2$ and $\displaystyle n$ terms.
. . Its sum is: .$\displaystyle 2\cdot\frac{2^n-1}{2-1}\:=\:2(2^n-1)$
Hence, [3] becomes: .$\displaystyle -S_n \;=\;2(2^n-1) - n\cdot2^{n+1} \quad\Rightarrow\quad -S_n \;=\;2^{n+1} - 2 - n\cdot2^{n+1}$
. . $\displaystyle -S_n \;=\;-2 - (n-1)2^{n+2} \quad\Rightarrow\quad \boxed{S_n \;=\;(n-1)2^{n+1} + 2}$