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Worked solution

Split mixed sigma sums into arithmetic and geometric parts

Find (i) ∑k=0n(k+n+1)\sum_{k=0}^n(k+n+1) in terms of nn; (ii) ∑r=1N(4r+2−r)\sum_{r=1}^N(4r+2^{-r}) in terms of NN; (iii) ∑k=130(a2k+ka)\sum_{k=1}^{30}(a^{2k}+ka) in terms of aa.

CJC · Discussion · Question 9

Answer

(i) (n+1)(3n+2)/2(n+1)(3n+2)/2. (ii) 2N(N+1)+1−2−N2N(N+1)+1-2^{-N}. (iii) a2(1−a60)/(1−a2)+465aa^2(1-a^{60})/(1-a^2)+465a for a≠±1a\ne\pm1; 495495 at a=1a=1, −435-435 at a=−1a=-1.

(i) Count the constant term n + 1 times

The sum is n(n+1)/2+(n+1)2=12(n+1)(3n+2)n(n+1)/2+(n+1)^2=\tfrac12(n+1)(3n+2).

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(ii) Add the arithmetic and geometric sums

4∑r=1Nr=2N(N+1)4\sum_{r=1}^Nr=2N(N+1) and ∑r=1N2−r=1−2−N\sum_{r=1}^N2^{-r}=1-2^{-N}. Thus the answer is 2N(N+1)+1−2−N2N(N+1)+1-2^{-N}.

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(iii) Check the exceptional geometric ratios

The sum is ∑k=130a2k+a∑k=130k\sum_{k=1}^{30}a^{2k}+a\sum_{k=1}^{30}k. For a2≠1a^2\ne1, this is a2(1−a60)/(1−a2)+465aa^2(1-a^{60})/(1-a^2)+465a, equivalently a2(a60−1)/(a2−1)+465aa^2(a^{60}-1)/(a^2-1)+465a. If a=1a=1, evaluate directly to get 30+465=49530+465=495. If a=−1a=-1, it is 30−465=−43530-465=-435.

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Common mistakes

Starting at k=0 gives n+1 terms; a geometric ratio of 1 needs a separate evaluation.

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