How to prove a binary decomposition of $x(1-x)$?

Finally I made a forehead proof.

$\sum\limits_{k=1}^{+\infty}2^{k-1}\left(-\frac{b_k}{2^k}+\sum\limits_{i=k+1}^{+\infty}\frac{b_i}{2^i}\right)^2 =\sum\limits_{k=1}^{+\infty}\left(b_k^2\frac{1}{2^{2k}}\sum\limits_{i=1}^{k}2^{i-1}\right) + \sum\limits_{i<j}\left(b_ib_j\left(\frac{2}{2^{i+j}}\sum\limits_{l=1}^{i-1}2^{l-1}-\frac{2^i}{2^{i+j}}\right)\right) = \sum\limits_{k=1}^{+\infty}\left(b_k^2\frac{1}{2^{2k}}(2^k-1)\right) + \sum\limits_{i<j}\left(b_ib_j\left(\frac{2}{2^{i+j}}(2^{i-1}-1)-\frac{2^i}{2^{i+j}}\right)\right) = -\sum\limits_{k=1}^{+\infty}\frac{b_k^2}{2^{2k}} + \sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k} - \sum\limits_{i<j}\frac{b_ib_j}{2^{i+j-1}} = \left(\sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k}\right)\left(1-\sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k}\right) = x(1-x)$

Where the trick was to use $b_k^2 = b_k$