# deriving formula for series?

Can any general formula (dependent on $n$) be derived for this expression:

$$\sum_{k = 1}^n 2^k k^2$$

If yes , then how we determine that any series can be converted into formula and what are ways to do that.

• I am new to this group and don't know how to add characters like Sigma and all , that's why wrote this expression like that. Please also tell me how you write such characters .
– user35921
Jul 17, 2012 at 21:57
• if you edit your question, you can see what the source format looks like. You can cancel the editing if you want after your curiosity is satisfied. Jul 17, 2012 at 22:05
• @tesla You should take a look at some of the links given in the answers to this question. Jul 17, 2012 at 22:06
• Jul 17, 2012 at 22:22

Start with : $$f_n(x)=\sum_{k=1}^n x^k$$ $$x f'_n(x)=\sum_{k=1}^n kx^k$$ $$x(x f'_n(x))'=\sum_{k=1}^n k^2x^k$$ What is $f_n(x)$?
Conclude...

Concerning the sentence "any series can be converted into formula". Some sophisticated methods exist but they don't work for 'any formula' (see Gosper algorithm, Zeilberger algorithm...).
A good starting point is to study generating functions for example in Wilf's excellent free book generatingfunctionology.

Hoping it helped,

• This is the way I'd go. (+1) Jul 17, 2012 at 23:43

I think so. (At first I thought it was $2^{-k}$, but $2^k$ will be more cumbersome.

It is the following:

$\sum_1^n 2^k + 3\sum_2^n 2^k + 5\sum_3^n 2^k+ \dots$

The summation in each case is $2\left(2^{r} - 1\right)$ where $r$ is the number of terms in the sum. The first sum has $n$ terms, second $n -1$ and so on.

Then, you have:

$2\left[2^n + 3.2^{n-1} + 5.2^{n -2} + \dots \right] - 2\left[1 + 3 + 5 + \dots\right]$

If you were considering, $\displaystyle\sum_1^\infty 2^{-k}.k^2$, you would get a pretty neat expression. But here, the common ratio is greater than 1, so it is a bit tough.

The answer may ultimately be NO. But, now you know what to do for common ratios less than 1.