Why is this expansion true?

Question

Why are this two sums equal?

$$ \sum_{i=1}^n i2^i = \sum_{j=1}^i\sum_{i=j}^n 2^i$$

I'm supposed to prove that both sums are the same.

Answer 1

HINT $$ \sum_{j=1}^k\sum_{i=j}^n 2^k = \sum_{j=1}^k 2^k \left(\sum_{i=j}^n 1 \right) $$

Answer 2

On the left

$\sum_{i=1}^n i\,x^i = x\dfrac{{\rm d}}{{\rm d}x}\sum_{i=1}^n x^i$

and on the right

$\sum_{i=j}^n x^i = \sum_{i=1}^n x^i - \sum_{i=1}^{j-1} x^i$

and using these re-writes you may reduce all expressions reduce via

$\sum_{k=1}^m x^k = x\,\dfrac{1-x^m}{1-x}$

Of course you may then substitute $x=2$, if you want.

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You may check it on worlframalpha or so via

Sum[i*x^i, {i, 1, n}] - Sum[Sum[x^i, {i, j, n}], {j, 1, n}] // Simplify

I did that here.

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It's not like the inner sum on the right hand side equals $i\,2^i$ so that there would be a straight forward reduction. Nevertheless, it's almost a given that there is a double counting argument and some sum that just reduces to $i\,2^i$ that "should" be written on the left hand side.

Answer 3

The r.h.s. should be $$\sum_{i=1}^n\sum_{j=1}^i 2^i,$$ and the equality is obvious.

Answer 4

$$\begin{align} \sum_{i=1}^n i 2^i &=\sum_{i=1}^n\sum_{j=1}^i 2^i\\ &=\sum_{1\le j\le i\le n} 2^i\\ &=\sum_{j=1}^\color{red}{n}\sum_{i=j}^n 2^i \end{align}$$

Answer 5

Without words:

$$1\cdot2^1+2\cdot2^2+3\cdot2^3+4\cdot2^4+\cdots\\=\\ 2^1+2^2+2^3+2^4+\cdots\\ \ \ \ \ \ \ \ \ \ 2^2+2^3+2^4+\cdots\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^3+2^4+\cdots\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^4+\cdots\\ $$