Sigma solving

**ukrobo** · May 23rd 2010, 06:08 AM

How do i simplify this in terms of n?

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

Cheers!

**galactus** · May 23rd 2010, 07:46 AM

Originally Posted by ukrobo

How do i simplify this in terms of n?

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

Cheers!

$\sum_{i=0}^{n-1}{2^{i} (n-i)}=\frac{1}{2}\sum_{i=1}^{n}2^{i}+\frac{n}{2}\sum _{i=1}^{n}2^{i}-\frac{1}{2}\sum_{i=1}^{n}i2^{i}$

Try using the partial sum for a geometric series formula.

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

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

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

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

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

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

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

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

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

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

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

Add this with the other two sums and we get:

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

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.

**Danny** · May 23rd 2010, 09:13 AM

Originally Posted by ukrobo

How do i simplify this in terms of n?

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

Cheers!

Here's another way. Let

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

multiplying by 2 gives

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

Now shift

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

Now substract (1) from (2) so

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

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