$\begin{equation*}\begin{array}{l}
1\cdot 2^{n+1} + 2 \cdot 2^n + \ldots + (n+1) \cdot 2 + (n+2) = \\
\textrm{izlu\v{c}i 2 iz prvih $n+1$ pribrojnika} \\
2 \cdot (1\cdot 2^n + 2 \cdot 2^{n-1} + \ldots + (n+1)) + (n+2) = \\
\textrm{po pretpostavci indukcije} \\
2 \cdot (2^{n+2} - (n - 3)) + (n+2) = \\
2^{n+3} - 2n + 6 + n + 2 = \\
2^{n+3} - n + 8 = \\
2^{(n+1)+2} - ((n+1) - 3) + 6
\end{array}\end{equation*}$