absolutestylehc

2022-01-21

Given
$1\cdot {2}^{1}+2\cdot {2}^{2}+3\cdot {2}^{3}+4\cdot {2}^{4}+\dots +d\cdot {2}^{d}=\sum _{r=1}^{d}r\cdot {2}^{r}$
how can we infer to the following solution?
$2\left(d-1\right)\cdot {2}^{d}+2$

Jordyn Horne

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

Deegan Mullen

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

RizerMix

Summation by parts gives that with the choice we have ${A}_{k}={2}^{1}+...+{2}^{k}={2}^{k+1}-2$ and $\sum _{k=1}^{n}{a}_{k}{b}_{k}={A}_{n}{b}_{n}-\sum _{k=1}^{n-1}{A}_{k}$ so: $\sum _{k=1}^{n}k{2}^{k}=\left({2}^{n+1}-2\right)n-\sum _{k=1}^{n-1}\left({2}^{k+1}-2\right)=\left({2}^{n+1}-2\right)n+2\left(n-1\right)-2\left({2}^{n}-2\right)$ and the RHS simplifies to $\left(n-1\right){2}^{n+1}+2$ as wanted.