$\sum_{k=1}^{100}k\cdot 2^k\geq 99\cdot 2^{101}$ solve geometrically?

Question

I've succeeded proving it using induction, but I wondered if there exists a prettier solution, maybe prooving it geometrically with areas of rectangles or some other way.

Answer 1

$$\sum_{k=1}^{100}k2^k=2\left(\sum_{k=1}^{100}x^k\right)'_{x=2}=2\left(\frac{x^{101}-x}{x-1}\right)'_{x=2}=2\cdot\frac{101\cdot2^{100}-1-2^{101}+2}{(2-1)^2}>99\cdot2^{101}.$$

Answer 2

Double and subtract, then repeat:

If $S=\sum\limits_{k=1}^{100}k\cdot 2^k$

then $2S=\sum\limits_{k=1}^{100}k\cdot 2^{k+1} =\sum\limits_{k=2}^{101}(k-1)\cdot 2^{k}$

so $S=2S-S = 100\cdot2^{101}-2-\sum\limits_{k=2}^{100}2^{k}$

and similarly $2S = 200\cdot2^{101}-4-\sum\limits_{k=3}^{101}2^{k}$

so now $S=2S-S = 100\cdot2^{101} -2 -2^{101}+2^2 = 99\cdot2^{101} +2 $

meaning $S \gt 99\cdot2^{101}$