1. 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

2. Re: 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?

3. Re: Proving a summation

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

And differentiate both sides with respect to $\displaystyle a$, you can also get the formula you want.

4. 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?

5. Re: 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:

$\displaystyle (1+a+a^2+\ldots+a^{n+1})(1-a)=$
$\displaystyle \begin{array}{cccccccccccc} & 1 & + & a & + & a^2 & + & \ldots & + & a^{n+1} & & \\ & & - & a & - & a^2 & - & \ldots & - & a^{n+1} & - & a^{n+2} \\ = & 1 & & & & & & & & & - & a^{n+2}\end{array}$

6. Re: 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?
$\displaystyle \frac{a}{(1 - a)^2}[1 - (n + 1)a^n + na^{n+1}]$

7. 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.