1970 AMC 12 第 22 题

Question

若前

3n

个正整数之和比前

n

个正整数之和多

150

，则前

4n

个正整数之和为：

If the sum of the first

3n

positive integers is

150

more than the sum of the first

n

positive integers, then the sum of the first

4n

positive integers is:

300

350

400

450

600

利用

1+2+\cdots+m=\frac{m(m+1)}{2}

Use

1+2+\cdots+m=\frac{m(m+1)}{2}

题目所给的差化简为

4n^2+n

The given difference simplifies to

4n^2+n

设

S_m=\frac{m(m+1)}{2}

。则

\begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n \end{aligned}\text{。}

由于

S_{3n}-S_n=150

，可得

4n^2+n=150

。因此

(n-6)(4n+25)=0

，其正数解为

n=6

。所以

S_{4n}=S_{24}=\frac{24\cdot25}{2}=300\text{。}

因此，正确答案是 A。

Let

S_m=\frac{m(m+1)}{2}.

Then

\begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n. \end{aligned}

Since

S_{3n}-S_n=150,

this gives

4n^2+n=150.

Hence

(n-6)(4n+25)=0,

and the positive solution is

n=6.

Therefore

S_{4n}=S_{24}=\frac{24\cdot25}{2}=300.

Therefore, the correct answer is A.

其他年份的第 22 题