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

absolutestylehc

absolutestylehc

Answered

2022-01-21

Given
121+222+323+424++d2d=r=1dr2r
how can we infer to the following solution?
2(d1)2d+2

Answer & Explanation

Jordyn Horne

Jordyn Horne

Expert

2022-01-22Added 16 answers

As noted above, observe that
r=1dxr=x(xd1)x1
Differentiating both sides and multiplying by x, we find
r=1drxr=dxd+2xd+1(d+1)+x(x1)2
Substituting x=2
r=1dr2r=d2d+2(d+1)2d+1+2=(d1)2d+1+2
Deegan Mullen

Deegan Mullen

Expert

2022-01-23Added 12 answers

You can use iterative integration of finite differences. The method works as follows. Assume your problem statement is:
r=1nar=?
Assume there is a function Sn such that:
r=1nar=SnS1+c
By finite difference we have:
an=r=1narr=1n1ar=(SnS1+c)(Sn1S1+c)
=SnSn1=Sn
Now assume we have a guess Tn for Sn such that some error Sn remains:
Sn=Tn+Sn
The error itself can be expressed as a new integral since finite difference distributes over sums. We have:
an=Sn=Tn+Sn
And hence we have a new sub problem:
r=1nar=r=1n(arTr)=SnS1+c
Lets apply the method to the problem at hand, we have:
ar=r2r
We can guess:
Tn=n2n+1
We arrive at:
ar=r2r(r2r+1(r1)2r1+1)=2r
We can guess again:
Tn=2n+1
We arrive at:
a=2r(2r+12r1+1)=0
So the integration terminated, and the closed form integral is:
Sn=Tn+Tn=n2n+12n+1
Using this for the sum we get:
r=1nr2r=SnS1+c=n2n+12n+1(121+121+1)+c
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Summation by parts gives that with the choice ak=2k, bk=k we have Ak=21+...+2k=2k+12 and k=1nakbk=Anbnk=1n1Ak so: k=1nk2k=(2n+12)nk=1n1(2k+12)=(2n+12)n+2(n1)2(2n2) and the RHS simplifies to (n1)2n+1+2 as wanted.

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