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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.

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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
Difficulty rating: 1630
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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