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

Question

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

Jordyn Horne · Accepted Answer

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

Deegan Mullen · Accepted Answer

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=Sn−S1+cBy finite difference we have:an=∑r=1nar−∑r=1n−1ar=(Sn−S1+c)−(Sn−1−S1+c)=Sn−Sn−1=△SnNow 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(ar−△Tr)=Sn′−S1′+cLets apply the method to the problem at hand, we have:ar=r⋅2rWe can guess:Tn=n⋅2n+1We arrive at:ar′=r⋅2r−(r⋅2r+1−(r−1)⋅2r−1+1)=−2rWe can guess again:Tn′=−2n+1We arrive at:a″=−2r−(−2r+1−2r−1+1)=0So the integration terminated, and the closed form integral is:Sn=Tn+Tn′=n⋅2n+1−2n+1Using this for the sum we get:∑r=1nr⋅2r=Sn−S1+c=n⋅2n+1−2n+1−(1⋅21+1−21+1)+c

RizerMix · Accepted Answer

Summation by parts gives that with the choice ak=2k, bk=k we have Ak=21+...+2k=2k+1−2 and ∑k=1nakbk=Anbn−∑k=1n−1Ak so: ∑k=1nk2k=(2n+1−2)n−∑k=1n−1(2k+1−2)=(2n+1−2)n+2(n−1)−2(2n−2) and the RHS simplifies to (n−1)2n+1+2 as wanted.

Given 1\cdot2^1+2\cdot2^2+3\cdot2^3+4\cdot2^4+...+d\cdot2^d=\sum_{r=1}^d r\cdot 2^r how can we infer to the following solution? 2(d-1)\cdot2^d+2

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