# Sigma solving

#### ukrobo

How do i simplify this in terms of n?

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

Cheers!

#### galactus

MHF Hall of Honor
How do i simplify this in terms of n?

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

Cheers!
$$\displaystyle \sum_{i=0}^{n-1}{2^{i} (n-i)}=\frac{1}{2}\sum_{i=1}^{n}2^{i}+\frac{n}{2}\sum_{i=1}^{n}2^{i}-\frac{1}{2}\sum_{i=1}^{n}i2^{i}$$

Try using the partial sum for a geometric series formula.

$$\displaystyle S_{n}=a_{n}\left(\frac{1-r^{n}}{1-r}\right)$$, where

r is the common ratio and a is the first term.

So, from the first one we have $$\displaystyle \frac{1-2^{n}}{1-2}=2^{n}-1$$

The second one is the same only multiplied by n. Giving $$\displaystyle n(2^{n}-1)$$

The third only can be found by differentiating the partial sum formula.

$$\displaystyle \sum_{i=1}^{n}x^{i}=\frac{x(1-x^{n})}{1-x}$$

$$\displaystyle \frac{d}{dx}\left[\sum_{i=1}^{n}x^{i}\right]=\sum_{i=1}^{n}ix^{i-1}=\frac{(nx-n-1)x^{n}+1}{(x-1)^{2}}$$

$$\displaystyle \sum_{i=1}^{n}ix^{i}=\frac{x((nx-x-1)x^{n}+1)}{(x-1)^{2}}$$

Sub in x=2 and we get $$\displaystyle 2(2^{n}(n-1)+1)$$.

Don't forget to divide by 2 because of the 1/2.

$$\displaystyle \frac{1}{2}\sum_{i=1}^{n}i2^{i}=(n-1)2^{n}+1$$

Add this with the other two sums and we get:

$$\displaystyle 2^{n}-1+n(2^{n}-1)-((n-1)2^{n}+1)=\boxed{2^{n+1}-n-2}$$

Just one way to go about it. Remember, you can play around with the geometric series formulas by differentiating, integrating, and so on, and get away with a lot.

#### Jester

MHF Helper
How do i simplify this in terms of n?

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

Cheers!
Here's another way. Let

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

multiplying by 2 gives

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

Now shift

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

Now substract (1) from (2) so

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