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.
 A: 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,
A: 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.
