$\begin{align}  & \sum\limits_{r=1}^{n}{(r+1)\cdot {{2}^{r}}}=n\cdot {{2}^{n+1}} \\ & \text{When }n=1 \\ & L.H.S(1+1)\cdot {{2}^{1}}=2\cdot {{2}^{1}}=4 \\ & R.H.S1\cdot {{2}^{1+1}}={{2}^{2}}=4 \\ & L.H.S=R.H.S \\ & \text{Thus the formula is valid when }n=1 \\ & \text{Assume the proposition is true for }n=k \\ & \sum\limits_{r=1}^{k}{(k+1)\cdot {{2}^{k}}}=k\cdot {{2}^{k+1}}\text{ and we wish to show that } \\ & {{S}_{k+1}}\text{ is also valid} \\ & {{S}_{k+1}}=(k+1)\cdot {{2}^{k+2}}---R.H.S \\ & \text{To do this } \\ & {{\text{S}}_{k+1}}={{S}_{k}}+{{a}_{k+1}}\text{ (By induction principle)} \\ & {{S}_{k+1}}=\sum\limits_{r=1}^{k}{(k+1)\cdot {{2}^{k}}}+(k+1+1)\cdot {{2}^{k+1}} \\ & {{S}_{k+1}}=k\cdot {{2}^{k+1}}+(k+2)\cdot {{2}^{k+1}} \\ & {{S}_{k+1}}={{2}^{k+1}}(k+k+2)={{2}^{k+1}}(2k+2) \\ & {{S}_{k+1}}={{2}^{1}}(k+1)\cdot {{2}^{k+1}}=(k+1)\cdot {{2}^{k+2}}---R.H.S \\ & \text{which shows that the statement is true for }n=k+1, \\ & \text{we can conclude that the given statement is true for } \\ & \text{all positive integers }n. \\\end{align}$