Sum of squares of natural numbers and geometric series

Question

$$S_n = 0\cdot4^0 + 1^2\cdot4^1 + 2^2\cdot4^2 + 3^2\cdot4^3 + \ldots + n^2\cdot4^n$$

ie- $S_n = \sum\limits_{i=0}^ni^2\cdot4^i$.

I need help with this problem. I integrated it using ILATE rule and found the leading term of this series of the order $O(n^2\cdot4^n)$. However, I want the exact sum of this series since I want the co-efficient of the leading term of this series.

I am working on a recurrence relation problem and after unfolding the recurrence relation, I got this series. However, I am not able to think of an approach to solve this series.

Hint: Sums like this can be addressed by (repeatedly) differentiating the geometric series $\sum_{i=0}^kx^i$ and adjusting the terms as needed. — lulu, Jan 30 '21 at 18:43
$f(x) := \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \implies f'(x) =\sum_{k=1}^n kx^{k-1} = \frac{d}{dx}\left(\frac{1-x^{n+1}}{1-x}\right)$. Multiply by $x$ then differentiate one more time. — Asem Abdelraouf, Jan 30 '21 at 18:48

score 1 · Accepted Answer · answered Jan 31 '21 at 06:44

1

Consider $$S_n(x) = \sum\limits_{i=0}^n i^2x^i= \sum\limits_{i=0}^n\big[ i(i-1)+i\big]x^i$$ $$S_n(x) =x^2 \sum\limits_{i=0}^n i(i-1)x^{i-2}+x\sum\limits_{i=0}^n ix^{i-1}$$ $$S_n(x) =x^2 \left(\sum\limits_{i=0}^n x^i \right)''+x\left(\sum\limits_{i=0}^n x^i \right)'$$ $$\sum\limits_{i=0}^n x^i=\frac{x^{n+1}-1}{x-1}$$ When done, let $x=4$.

answered Jan 31 '21 at 06:44

Claude Leibovici

289,558

Much more elegant than the approach that I originally took in this answer. I ended up stealing your approach. – user2661923 Jun 16 '24 at 03:08

Sum of squares of natural numbers and geometric series

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