# Proving a summation

• Oct 15th 2013, 08:10 AM
mohamedennahdi
Proving a summation
Dear all, how can I prove the following summation (See attachment)?
I tried to use integrals but the result was kind of broad.

Attachment 29481
• Oct 15th 2013, 08:16 AM
SlipEternal
Re: Proving a summation
$0 = \sum_{k = 0}^0 ka^k = \dfrac{a}{(1-a)^2}[1-(0+1)a^0 + 0a^{0+1}] = 0$

Assume the formula is correct for all nonnegative integers up to $n$.

$\sum_{k=0}^{n+1}ka^k = \sum_{k=0}^n ka^k + (n+1)a^{n+1}$

By the induction assumption, we have

$\sum_{k=0}^n ka^k + (n+1)a^{n+1} = \dfrac{a}{(1-a)^2}[1-(n+1)a^n + na^{n+1}] + (n+1)a^{n+1}$

Can you finish from here?
• Oct 15th 2013, 08:23 AM
SlipEternal
Re: Proving a summation

$\dfrac{1-a^{n+2}}{1-a} = \sum_{k=0}^{n+1}a^k$

And differentiate both sides with respect to $a$, you can also get the formula you want.
• Oct 15th 2013, 08:34 AM
mohamedennahdi
Re: Proving a summation
Thank you.
abd Could you help me with how we can come up with the formula from the summation?
As how to deduce the formula from the sigma notation?
• Oct 15th 2013, 08:50 AM
SlipEternal
Re: Proving a summation
The original formula? Or the formula I wrote: $\dfrac{1-a^{n+2}}{1-a} = \sum_{k = 0}^{n+1}a^k$? Because that formula is just the standard geometric sum:

$(1+a+a^2+\ldots+a^{n+1})(1-a)=$
$\begin{array}{cccccccccccc} & 1 & + & a & + & a^2 & + & \ldots & + & a^{n+1} & & \\ & & - & a & - & a^2 & - & \ldots & - & a^{n+1} & - & a^{n+2} \\ = & 1 & & & & & & & & & - & a^{n+2}\end{array}$
• Oct 15th 2013, 12:35 PM
mohamedennahdi
Re: Proving a summation
Imagine you have only the sigma notation, like this:
$\sum_{i = 0}^{n}i2^i = ?$
How would you deduce the following formula from the above summation?
$\frac{a}{(1 - a)^2}[1 - (n + 1)a^n + na^{n+1}]$
• Oct 15th 2013, 01:11 PM
SlipEternal
Re: Proving a summation
I would think it looks like the derivative of a geometric sum. So, I would start with a geometric sum, and take the derivative to see what I get.