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

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- Oct 15th 2013, 07:10 AMmohamedennahdiProving 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, 07:16 AMSlipEternalRe: Proving a summation
$\displaystyle 0 = \sum_{k = 0}^0 ka^k = \dfrac{a}{(1-a)^2}[1-(0+1)a^0 + 0a^{0+1}] = 0$

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

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

By the induction assumption, we have

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

Can you finish from here? - Oct 15th 2013, 07:23 AMSlipEternalRe: Proving a summation
Alternately, if you start with

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

And differentiate both sides with respect to $\displaystyle a$, you can also get the formula you want. - Oct 15th 2013, 07:34 AMmohamedennahdiRe: 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, 07:50 AMSlipEternalRe: Proving a summation
The original formula? Or the formula I wrote: $\displaystyle \dfrac{1-a^{n+2}}{1-a} = \sum_{k = 0}^{n+1}a^k$? Because that formula is just the standard geometric sum:

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

$\displaystyle \begin{array}{cccccccccccc} & 1 & + & a & + & a^2 & + & \ldots & + & a^{n+1} & & \\ & & - & a & - & a^2 & - & \ldots & - & a^{n+1} & - & a^{n+2} \\ = & 1 & & & & & & & & & - & a^{n+2}\end{array}$ - Oct 15th 2013, 11:35 AMmohamedennahdiRe: Proving a summation
Imagine you have only the sigma notation, like this:

$\displaystyle \sum_{i = 0}^{n}i2^i = ?$

How would you deduce the following formula from the above summation?

$\displaystyle \frac{a}{(1 - a)^2}[1 - (n + 1)a^n + na^{n+1}]$ - Oct 15th 2013, 12:11 PMSlipEternalRe: 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.