# 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 '12 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. – hmakholm left over Monica Jul 17 '12 at 22:05
• @tesla You should take a look at some of the links given in the answers to this question. – Adrián Barquero Jul 17 '12 at 22:06
• – joriki Jul 17 '12 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) – user26872 Jul 17 '12 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.