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Question: Evaluate

$$ \sum_{i=0}^n i^2 2^{n-i}$$

In the previous questions on my question paper I have got:

$$\sum_{i=1}^n a^i = \frac{a(1-a^n)}{1-a}$$ and $$\sum_{i=1}^nia^i=\frac{a(1-a^n)}{1-a}-na^{n+1}$$

My step for this evaluation is below:

Let $ 2^{n-1} \rightarrow a^{n-i}$

$$\therefore \sum_{i=1}^ni^2a^{n-i}$$ $$a^n\sum_{i=1}^ni^2a^{-i}$$

Let $b = {1 \over a}$

$$\therefore S_n=({1 \over b})^n\sum_{i=1}^ni^2b^i$$

I am trying to use the "multiply b by both sides" method and eliminate $S_n-bS_n$

The result is like this:

$bS_n-S_n=(b-1)S_n=(b^2+2^2b^3+3^2b^4+\cdots+n^2b^{n+1})-(b+2^2b^2+3^2b^3+\cdots+n^2b^n)$

Which is

$$b+(2^2-1)b^2+(3^2-2^2)b^3+\cdots+(n^2-(n-1)^2)b^n-n^2b^{n+1}$$ Expand: $$b+2^2b^2-b^2+3^2b^3-2^2b^3+\cdots+n^2b^n-(n-1)^2b^n-n^2b^{n+1}$$

I am now stuck at this step. Did I do anything wrong?

PS: I heard that we could use the answers in the previous questions but I couldn't see the patterns. Would anyone help? Thanks.

UserX
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hlx98007
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1 Answers1

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Here is a convenient trick: if $|x| < 1$ you have the finite geometric series $$\sum_{k=0}^n x^k = \frac{x^{n+1} - 1}{x-1}.$$ Then $$\sum_{k=1}^n k x^k = x \sum_{k=0}^n k x^{k-1} = x \cdot \frac{d}{dx} \left( \frac{x^{n+1} - 1}{x-1} \right).$$ You can repeat this argument (after computing the derivative explicitly) to get a closed form for $\displaystyle \sum_{k=1}^n k^2 x^k$. Then observe $$\sum_{i=0}^n i^2 2^{n-i} = 2^n \sum_{i=1}^n i^2 \left( \frac 12 \right)^i.$$

Umberto P.
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  • Hi, I am not too sure about why $$x\sum_{k=0}^nkx^{k-1}$$ becomes $$x{d \over dx}({x^{n+1}-1 \over x-1})$$could you expand the process? – hlx98007 Sep 09 '14 at 14:31
  • $$ \sum_{k=0}^n kx^{k-1} = \sum_{k=0}^n (x^k)' = \left( \sum_{k=0}^n x^k \right)' = \left( \dfrac{x^{n+1} - 1}{x-1} \right)'$$ – Umberto P. Sep 09 '14 at 14:56
  • Sorry, I still don't see how $\displaystyle 2^n\sum_{i=1}^ni^2 ({1 \over 2})^i$ can be used in any way. – hlx98007 Sep 09 '14 at 18:41