The sum k=1∑∞k3+6k2+8k1 can be expressed as ba, where a and b are relatively prime positive integers. What is a+b?
Answer Choices:
A. 89
B. 97
C. 102
D. 107
E. 129
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By partial fraction decomposition, we have k3+6k2+8k1=81(k1−k+22+k+41), so we may write:
8k=1∑∞k3+6k2+8k1=k=1∑∞(k1−k+22+k+41)=k=1∑∞(k1−k+21)+k=1∑∞(k+41−k+21)=k=1∑∞(k1−k+21)−k=1∑∞(k+21−k+41)=k=1∑∞(k1−k+21)−k=3∑∞(k1−k+21)=k=1∑2(k1−k+21)
Alternatively, we may note the individual sums themselves telescope and compute those sums first. Either method gives an answer of 81(11+21−31−41)=81⋅1211=9611 which gives a sum of 11+96=(D) 107.
The problems on this page are the property of the MAA's American Mathematics Competitions