# How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$? [duplicate]

Possible Duplicate:
Formula for calculating $\sum_{n=0}^{m}nr^n$

Can someone explain step by step how to derive the following identity? $$\sum_{k=1}^{n} k \cdot 2^k = 2(n \cdot 2^n - 2^n + 1)$$

If you know $$\sum_{k=1}^n a^k=\frac{a^{n+1}-a}{a-1}$$, take the derivative of both sides with respect to $$a$$, multiply by $$a$$, and set $$a=2$$

• Since the lower index of the summation is $1$, the RHS should be $\frac{a^{n + 1} - \color{red}{a}}{a - 1}$. – N. F. Taussig Jan 23 '18 at 12:41

Although you got great answers, I will add another way to look at the expression you are looking for

Let us denote

$$S = \sum_{k=1}^{n} k \cdot 2^k$$

This can be expanded as

\begin{align*} S &= 2 + 2 . 2^2 + 3 . 2^3 + \cdots n 2^n\\ 2S &= \hspace{8pt}+1.2^2+2.2^3 + \cdots (n-1) . 2^n+n . 2^{n+1} \end{align*}

If you notice, by multiplying by $2$ and writing under similar terms and then subtracting, we get

\begin{align*} S = n . 2^{n+1} - \left(2+2^2+2^3+\cdots+2^n \right) &= n .2^{n+1}-2^{n+1}+2\\ &= 2(n . 2^n-2^n+1) \end{align*}

Induction is a surefire way to prove it with elementary methods. With calculus and some formulas, we can approach this problem by evaluating a derivative in two different ways, as follows:

Quotient rule: $$\rm \frac{d}{dx}\frac{x^n-1}{x-1} =\frac{n\,x^{n-1}(x-1)-(x^n-1)(1)}{(x-1)^2} \tag{1}$$

Geometric formula:

$$\rm \frac{d}{dx}\frac{x^n-1}{x-1}=0+1x^0+2x^1+\cdots+n\,x^{n-1} \tag{2}$$

Equate equation $(1)$ with equation $(2)$, multiply both sides by $\rm x$ and then set $\rm x=2$.

• Your last term in $(2)$ should be $nx^{n-1}$ instead – Kirthi Raman Mar 14 '12 at 16:07