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?

$2(d-1)\cdot {2}^{d}+2$

how can we infer to the following solution?

Jordyn Horne

Beginner2022-01-22Added 16 answers

As noted above, observe that

$\sum _{r=1}^{d}{x}^{r}=\frac{x({x}^{d}-1)}{x-1}$

Differentiating both sides and multiplying by x, we find

$\sum _{r=1}^{d}r{x}^{r}=\frac{{dx}^{d+2}-{x}^{d+1}(d+1)+x}{{(x-1)}^{2}}$

Substituting$x=2$

$\sum _{r=1}^{d}r{2}^{r}=d{2}^{d+2}-(d+1){2}^{d+1}+2=(d-1){2}^{d+1}+2$

Differentiating both sides and multiplying by x, we find

Substituting

Deegan Mullen

Beginner2022-01-23Added 12 answers

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:

${a}_{n}=\sum _{r=1}^{n}{a}_{r}-\sum _{r=1}^{n-1}{a}_{r}=({S}_{n}-{S}_{1}+c)-({S}_{n-1}-{S}_{1}+c)$

$={S}_{n}-{S}_{n-1}=\mathrm{\u25b3}{S}_{n}$

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:

$a}_{n}=\mathrm{\u25b3}{S}_{n}=\mathrm{\u25b3}{T}_{n}+\mathrm{\u25b3}{S}_{n}^{\prime$

And hence we have a new sub problem:

$\sum _{r=1}^{n}{a}_{r}^{\prime}=\sum _{r=1}^{n}({a}_{r}-\mathrm{\u25b3}{T}_{r})={S}_{n}^{\prime}-{S}_{1}^{\prime}+c$

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:

$a}_{r}^{\prime}=r\cdot {2}^{r}-(r\cdot {2}^{r+1}-(r-1)\cdot {2}^{r-1+1})=-{2}^{r$

We can guess again:

$T}_{n}^{\prime}=-{2}^{n+1$

We arrive at:

$a{}^{\u2033}=-{2}^{r}-(-{2}^{r+1}-{2}^{r-1+1})=0$

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:

$\sum _{r=1}^{n}r\cdot {2}^{r}={S}_{n}-{S}_{1}+c=n\cdot {2}^{n+1}-{2}^{n+1}-(1\cdot {2}^{1+1}-{2}^{1+1})+c$

Assume there is a function

By finite difference we have:

Now assume we have a guess

The error itself can be expressed as a new integral since finite difference distributes over sums. We have:

And hence we have a new sub problem:

Lets apply the method to the problem at hand, we have:

We can guess:

We arrive at:

We can guess again:

We arrive at:

So the integration terminated, and the closed form integral is:

Using this for the sum we get:

RizerMix

Expert2022-01-27Added 656 answers

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

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