Given 1⋅21+2⋅22+3⋅23+4⋅24+…+d⋅2d=∑r=1dr⋅2r how can we infer to the following solution? 2(d−1)⋅2d+2

As noted above, observe that
${\displaystyle \sum _{r=1}^d{x}^r=\frac{x(x^d-1)}{x-1}}$
Differentiating both sides and multiplying by x, we find
${\displaystyle \sum _{r=1}^{d}r{x}^r=\frac{{dx}^{d+2}-x^{d+1}(d+1)+x}{{(x-1)}^2}}$
Substituting ${\displaystyle x=2}$
${\displaystyle \sum _{r=1}^{d}r{2}^r=d{2}^{d+2}-(d+1)2^{d+1}+2=(d-1)2^{d+1}+2}$

You can use iterative integration of finite differences. The method works as follows. Assume your problem statement is:
${\displaystyle \sum _{r=1}^n{a}_r=?}$
Assume there is a function ${\displaystyle S_n}$ such that:
${\displaystyle \sum _{r=1}^n{a}_r=S_n-S_1+c}$
By finite difference we have:
${\displaystyle a_n=\sum _{r=1}^n{a}_r-\sum _{r=1}^{n-1}{a}_r=(S_n-S_1+c)-(S_{n-1}-S_1+c)}$
${\displaystyle =S_n-S_{n-1}=\mathrm{△}{S}_n}$
Now assume we have a guess ${\displaystyle T_n}$ for ${\displaystyle S_n}$ such that some error ${\displaystyle S_n^{\prime }}$ remains:
${\displaystyle S_n=T_n+S_n^{\prime }}$
The error itself can be expressed as a new integral since finite difference distributes over sums. We have:
${\displaystyle a_n=\mathrm{△}{S}_n=\mathrm{△}{T}_n+\mathrm{△}{S}_n^{\prime }}$
And hence we have a new sub problem:
${\displaystyle \sum _{r=1}^n{a}_r^{\prime }=\sum _{r=1}^n(a_r-\mathrm{△}{T}_r)=S_n^{\prime }-S_1^{\prime }+c}$
Lets apply the method to the problem at hand, we have:
${\displaystyle a_r=r\cdot 2^r}$
We can guess:
${\displaystyle T_n=n\cdot 2^{n+1}}$
We arrive at:
${\displaystyle a_r^{\prime }=r\cdot 2^r-(r\cdot 2^{r+1}-(r-1)\cdot 2^{r-1+1})=-2^r}$
We can guess again:
${\displaystyle T_n^{\prime }=-2^{n+1}}$
We arrive at:
${\displaystyle a{}^{″}=-2^r-(-2^{r+1}-2^{r-1+1})=0}$
So the integration terminated, and the closed form integral is:
${\displaystyle S_n=T_n+T_n^{\prime }=n\cdot 2^{n+1}-2^{n+1}}$
Using this for the sum we get:
${\displaystyle \sum _{r=1}^{n}r\cdot 2^r=S_n-S_1+c=n\cdot 2^{n+1}-2^{n+1}-(1\cdot 2^{1+1}-2^{1+1})+c}$

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