1958 AMC 12 Problem 37

Attempt Problem 37 of the 1958 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1958 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

37.

The first term of an arithmetic series of consecutive integers is k2+1.k^2+1. The sum of 2k+12k+1 terms of this series may be expressed as:
k3+(k+1)3k^3+(k+1)^3
(k−1)3+k3(k-1)^3+k^3
(k+1)3(k+1)^3
(k+1)2(k+1)^2
(2k+1)(k+1)2(2k+1)(k+1)^2
Answer: A
Concepts:arithmetic sequencesummationalgebraic manipulation
Small hint:
Find the last term after 2k2k increases from the first term
Big hint:
Use the arithmetic-series average of the first and last terms
Solution:
The last term is k2+1+2k=(k+1)2. k^2+1+2k=(k+1)^2. The average of the first and last terms is k2+k+1.k^2+k+1. Therefore the sum is S=(2k+1)(k2+k+1)=k3+(k+1)3. \begin{aligned} S&=(2k+1)(k^2+k+1)\\ &=k^3+(k+1)^3. \end{aligned}
Therefore, the correct answer is A.
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