What is the exact closed form of $2^1\cdot1^{2}+2^2\cdot2^{2}+2^3\cdot3^{2}+\cdots+2^n\cdot n^{2}$?

Question

Let $S(n)=$ $2^1\cdot1^{2}+2^2\cdot2^{2}+2^3\cdot3^{2}+\cdots+2^n\cdot n^{2}$

I can't guess the form of $S(n)$

However, I can find the exact closed form of

$2^1\cdot1+2^2\cdot2+2^3\cdot3+\cdots+2^n\cdot n$

which is $(n-1)2^{n+1}-2$

Answer 1

$S = \sum\limits_{i=1}^{n} 2^i . i^2$

$2S = \sum\limits_{i=1}^{n} 2^{i+1} . i^2$

Subtract first equation from second,

$S = 2^{1}.1^2 + [\sum\limits_{i=2}^{i=n}2^i . (2i-1)] - 2^{n+1}n^2$

$S = 2^{1}.1^2 + [2\sum\limits_{i=2}^{i=n}2^i . i] - [\sum\limits_{i=2}^{i=n}2^i] - 2^{n+1}n^2$

I think you can proceed further. Let me know if you want me to complete.

Answer 2

Here is another answer based on the comment by @Yuriy. Let

$S = \sum\limits_{i=1}^{n} x^n = \frac{x(1-x^{n})}{1-x}$

Now take the derivative of LHS wrt $x$.

$\frac{dS}{dx} = \sum\limits_{i=1}^{n}nx^{n-1}$

Now, multiply by $x$ and again differentiate wrt $x$.

$\frac{d(x\frac{dS}{dx})}{dx} = \sum\limits_{i=1}^{n} n^2 x^{n-1}$

Finally, multiply by $x$ and replace $x$ by $2$.

$(x\frac{d(x\frac{dS}{dx})}{dx})_{\,_{x=2}} = \sum\limits_{i=1}^{n} n^2 2^{n}$, which is the desired sum.

Now, compute LHS in above equation by using the closed form of $S$ from first equation. Do you want me to complete this?