How to Solve Summation by Hand

Question

I am having trouble solving the following summation by hand.

$$ \sum_{i=0}^{n-1} 2^i (n-i) $$

Can someone guide me in the right direction, especially for the $i \cdot 2^i$ part?

Answer 1

Let $s(n) =\sum_{i=0}^{n-1} 2^i(n-i) $.

I'm going to try to go from $n$ to $n+1$ and see what happens.

$\begin{array}\\ s(n+1)-s(n) &=\sum_{i=0}^{n} 2^i(n+1-i)-\sum_{i=0}^{n-1} 2^i(n-i)\\ &=\sum_{i=0}^{n} 2^i(n-i)+\sum_{i=0}^{n} 2^i-\sum_{i=0}^{n-1} 2^i(n-i)\\ &=\sum_{i=0}^{n-1} 2^i(n-i)+(2^{n+1}-1)-\sum_{i=0}^{n-1} 2^i(n-i)\\ &=2^{n+1}-1\\ \text{so that}\\ s(m)-s(0) &=\sum_{n=0}^{m-1} (s(n+1)-s(n))\\ &=\sum_{n=0}^{m-1} (2^{n+1}-1)\\ &=\sum_{n=1}^{m} 2^{n}-m\\ &=2^{m+1}-2-m\\ \text{so that}\\ s(m) &=s(0)+2^{m+1}-2-m\\ &=2^{m+1}-2-m \qquad\text{since } s(0) = 0\\ \end{array} $

It worked!

Answer 2

$$ (i+1)2^{i+1} - i 2^i = i 2^i + 2^{i+1} $$ $$ \sum_{i=0}^{n-1} i 2^i = \sum_{i=0}^{n-1}[(i+1)2^{i+1} - i 2^i] - \sum_{i=0}^{n-1}2^{i+1} = n2^n-2^{n+1}+2 $$

Answer 3

Alternatively, one can use generating function method to find the sum: We know that

$$ \sum_{n=0}^{\infty} 2^n x^n = \frac{1}{1-2x} \qquad \text{and} \qquad \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2}. $$

Thus it follows that

\begin{align*} \sum_{n=0}^{\infty} \bigg( \sum_{i=0}^{n} 2^i (n-i) \bigg) x^n &= \frac{1}{1-2x} \cdot \frac{x}{(1-x)^2} \\ &= \frac{2}{1-2x} - \frac{1}{1-x} - \frac{1}{(1-x)^2} \\ &= \sum_{n=0}^{\infty} ( 2^{n+1} - 1 - (n+1) ) x^n. \end{align*}

Therefore

$$ \sum_{i=0}^{n} 2^i (n-i) = 2^{n+1} - n - 2. $$

Answer 4

$\sum 2^i$ is a geometric series.

\begin{align} \sum_{i=0}^{n-1} i2^i&=2\sum_{i=1}^{n-1} i2^{i-1}\\&=2\sum_{i=1}^{n-1} \frac{d}{dx}\left(x^i\right)|_{x=2}\\&=2\frac{d}{dx}\sum_{i=1}^{n-1} \left(x^i\right)|_{x=2}\\ &=2\frac{d}{dx}\left(\frac{x(x^{n-1}-1)}{x-1}\right)|_{x=2} \\&=2\frac{d}{dx}\left(\frac{x^{n}-x}{x-1}\right)|_{x=2} \\&=2\frac{(x-1)(nx^{n-1}-1)-(x^n-x)}{(x-1)^2}|_{x=2} \\ &=2(n2^{n-1}-1-2^n+2)\\ &=2(n2^{n-1}-2^n+1) \end{align}

Answer 5

Here's a hint:

$$ \sum_{i=0}^k i 2^i = 2 + 2*2^2+ 3*2^3 + \dots + k* 2^k \\ = ( 2 + 2^2 + 2^3 + \dots + 2^k ) \\ + (2^2+2^3+\dots + 2^k) \\ + \dots + (2^{k-1}+2^k) + (2^k) \\ = 2(1+ \dots + 2^{k-1}) + \dots + 2^{k-1}(1+2) + 2^k \\ = 2(2^k-1)+2^2(2^{k-1}-1) + \dots + 2^{k-1}(2^2-1) \\ = (k-1) 2^{k+1} + 2^k - 2(1+2+\dots+2^{k-2}) \\ = (k-1) 2^{k+1} + 2.$$

Can you get it from here?

Answer 6

Making the problem more general, consider $$A_n=\sum_{i=0}^{n-1} x^i (n-i)=n\sum_{i=0}^{n-1} x^i -\sum_{i=0}^{n-1} ix^i= n\sum_{i=0}^{n-1} x^i-x\sum_{i=0}^{n-1} ix^{i-1}$$ that is to say $$A_n=n\left(\sum_{i=0}^{n-1} x^i \right)-x\left(\sum_{i=0}^{n-1} x^i \right)'$$ $$\sum_{i=0}^{n-1} x^i= \frac{x^n-1}{x-1}$$ $$\left(\sum_{i=0}^{n-1} x^i \right)'=\frac{n x^{n-1}}{x-1}-\frac{x^n-1}{(x-1)^2}$$ which make $$A_n=\frac{x \left(x^n-1\right)-n x+n}{(x-1)^2}$$ which you could use for any value of $x$.

Making $x=2$ will give $A_n=2^{n+1}-n-2$.