Sum of the sereis

**skyskiers** · September 15th 2008, 09:16 AM

Helllo, frnds,

Please help me in solving this series.

S(n) = (1/2) + (2^2)/2^2 + (3^2)/2^3 + (4^2)/2^4 + .........

So, we can write this series as

Sum of T(n) = [(n^2)/(2^n)] ...

Any ideas ..please help..

Thanx..

**Plato** · September 15th 2008, 09:57 AM

The placement of this question in an elementary mathematics forum is puzzling.
This requires a bit of calculus at a higher level to solve.
Questions: Do you simply want a answer for some simple reason?
Or are you required to supply reasons for the particular answer?

Please respond so that we know how to answer and to what extent the answer should be.

**skyskiers** · September 15th 2008, 10:06 AM

Hello,

I really apologize for the wrong placement of this question. I want the full answer with reasons. Shall I open a new thread in the relevant section?

Thanks

**ThePerfectHacker** · September 15th 2008, 10:30 AM

Hint 1: Evaluate $\sum_{n=1}^{\infty} n^2 x^n$ by differenciating term-by-term the expression $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ twice.

Hint 2: Now consider $\sum_{n=1}^{\infty} \frac{n^2x^n}{2^n} = \sum_{n=1}^{\infty} n^2 \left( \frac{x}{2} \right)^n$ and evaluate this sum using #1.

Hint 3: Let $x=1$ to get your answer.

**Plato** · September 15th 2008, 10:59 AM

$\begin{array}{l}<br /> f(r) = \sum\limits_{n = 1}^\infty {r^n } = \frac{r}{{1 - r}},\quad \left| r \right| < 1 \\ <br /> g(r) = rf'(r) = \sum\limits_{n = 1}^\infty {nr^n } = \frac{r}{{\left( {1 - r} \right)^2 }} \\ <br /> rg'(r) = \sum\limits_{n = 1}^\infty {n^2 r^n } = \color{red}{\frac{{r\left( {1 + r} \right)}}{{\left( {1 - r} \right)^3 }}} \\ <br /> \end{array}$
Now let $r=\frac{1}{2}$

**skyskiers** · September 15th 2008, 04:15 PM

Thanx a lot for your help!!!

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