find the sum

**rcmango** · February 8th 2007, 08:27 AM

find the sum of the given series:

heres the equation: http://img19.imageshack.us/img19/559/untitledyc9.jpg

I was told it could be done by first finding a and b using that 2n +1 = a(n+1) + b

and given as a hint that E(n + 1)x^n is the derivateive of a known power series already.

please help me with this.
thankyou.

**ThePerfectHacker** · February 8th 2007, 09:59 AM

Find the interval of absolute convergence and then.
$\sum_{n=0}^{\infty}(2n+1)x^n$
Expand,
$1+3x+5x^2+7x^3+...$
$(2x+4x^2+6x^3+...)+(1+x+x^2+x^3+...)$
$2x(1+2x+3x^2+...)+(1+x+x^2+...)$ . (*)
Okay, now we notice a very important thing.

First we know that,
$1+x+x^2+x^3+...=\frac{1}{1-x}$
Then, take derivative,
$1+2x^2+3x^2+...=\frac{x}{(1-x)^2}$ .
Substitute that into (*),
$2x\cdot \frac{x}{(1-x)^2}+\frac{1}{1-x}$
$\frac{2x^2}{(1-x)^2}+\frac{1-x}{(1-x)^2}$
$\frac{2x^2-x+1}{(1-x)^2}$

**rcmango** · February 8th 2007, 10:26 AM

thanks for the help,

but where does (2x + 4x^2 + 6x^3 +...) + (1 +...

come from?

also, what is the sum?

thankyou.

**ThePerfectHacker** · February 8th 2007, 12:02 PM

Originally Posted by rcmango

thanks for the help,

but where does (2x + 4x^2 + 6x^3 +...) + (1 +...

come from?

also, what is the sum?

thankyou.

Well, add them you get the same thing. No?

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