1958 AMC 12 第 37 题

Question

一个由连续整数组成的等差数列首项为

k^2+1

。该数列前

2k+1

项的和可表示为：

The first term of an arithmetic series of consecutive integers is

k^2+1.

The sum of

2k+1

terms of this series may be expressed as:

k^3+(k+1)^3

(k-1)^3+k^3

(k+1)^3

(k+1)^2

(2k+1)(k+1)^2

从首项起增加

2k

次，求出末项

Find the last term after

2k

increases from the first term

利用等差数列首末项的平均数

Use the arithmetic-series average of the first and last terms

末项为

k^2+1+2k=(k+1)^2\text{。}

首项与末项的平均数为

k^2+k+1

。因此总和为

\begin{aligned} S&=(2k+1)(k^2+k+1)\\ &=k^3+(k+1)^3 \end{aligned}\text{。}

所以正确答案为 A。

The last term is

k^2+1+2k=(k+1)^2.

The average of the first and last terms is

k^2+k+1.

Therefore the sum is

\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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