So you're on the n=k+1 step. Alright.
Start with this
For the n=k+1 case
Notice that for the n=k case
![](https://latex.codecogs.com/png.latex?\bg_white \sum_{r=1}^{k}\left(r+1\right)\times{2^{r}}=k\times{2^{k+1}})
.
Thus, for the n=k+1 case
Working with the LHS
![](https://latex.codecogs.com/png.latex?\bg_white \sum_{r=1}^{k+1}\left(r+1\right)\times{2^{r}})
is just
![](https://latex.codecogs.com/png.latex?\bg_white k\times{2^{k+1}}+\left(k+2\right)\times{2^{k+1}}=k\cdot{2^{k+2}}+2^{k+2})
using the assumption made from the n=k statement
Cleaning this up on the LHS we will have
![](https://latex.codecogs.com/png.latex?\bg_white k\cdot{2^{k+1}}+k\cdot{2^{k+1}}+2^{k+2})
That simply gives us
![](https://latex.codecogs.com/png.latex?\bg_white k\cdot{2^{k+2}}+2^{k+2})
equalling the RHS.
The rest is up to you. But I recommend to write your induction proof as if you are completing an essay according to Bill Pender.