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.

  • 1
    $\begingroup$ 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 . $\endgroup$ – user35921 Jul 17 '12 at 21:57
  • 1
    $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Jul 17 '12 at 22:05
  • $\begingroup$ @tesla You should take a look at some of the links given in the answers to this question. $\endgroup$ – Adrián Barquero Jul 17 '12 at 22:06
  • $\begingroup$ See math.stackexchange.com/questions/30732/… and math.stackexchange.com/questions/50919/…. $\endgroup$ – 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 $f_n(x)$?

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,

  • $\begingroup$ This is the way I'd go. (+1) $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy