If $n\ge 0$ is an integer, then $\sum_{j=0}^nj\cdot 2^j=(n-1)\cdot 2^{n+1}+2$
We try it out for $n=0\text{ and } 1$. We suppose that $\sum_{j=0}^nj\cdot 2^j=(n-1)\cdot 2^{n+1}+2$ and prove for $n+1$:
$\sum_{j=0}^{n+1}j\cdot 2^j=n\cdot 2^{n+2}+2\leftrightarrow \sum_{j=0}^nj\cdot 2^j+(n+1)\cdot 2^{n+1}=n\cdot 2^{n+1}\cdot 2+2=2^{n+1}\cdot\left(2n\right)+2$ which is by induction: $\left((n-1)\cdot 2^{n+1}+2\right)+(n+1)\cdot 2^{n+1}=2^{n+1}\cdot\left(2n\right)+2\leftrightarr ow2^{n+1}\left(n-1+n+1\right)=2^{n+1}\cdot\left(2n\right)$ QED.