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Sissa and arithmetico-geometric sums

Authors
  • avatar
    Name
    Vu Hung
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Problem Statement

  • Evaluate S64=1+2+4+⋯+263S_{64}=1+2+4+\cdots+2^{63}.
  • Let
T=∑n=164n2n−1.T=\sum_{n=1}^{64} n2^{n-1}.

Use shift-and-subtract to find a closed form.

  • Evaluate
∑n=1∞n2n.\sum_{n=1}^{\infty}\frac{n}{2^n}.

Hints

Use T−2TT-2T for the finite AGP, and similarly S−12SS-\frac12S for the infinite one.

Solutions

S64=264−1.S_{64}=2^{64}-1.

Also

T−2T=∑n=1642n−1−64⋅264=(264−1)−64⋅264,T-2T=\sum_{n=1}^{64}2^{n-1}-64\cdot2^{64} =(2^{64}-1)-64\cdot2^{64},

so

T=63⋅264+1.T=63\cdot2^{64}+1.

For S=∑n=1∞n2nS=\sum_{n=1}^\infty \frac{n}{2^n}:

S−12S=12+14+18+⋯=1,S-\frac12S=\frac12+\frac14+\frac18+\cdots=1,

hence S=2S=2.

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