1979 AMC 12 第 14 题

Question

在某个数列中，第一项是

1

，并且对于所有

n\ge2

，前

n

项的乘积为

n^2

。该数列的第三项与第五项之和为

In a certain sequence of numbers, the first number is

1,

and, for all

n\ge2,

the product of the first

n

numbers in the sequence is

n^2.

The sum of the third and the fifth numbers in the sequence is

\frac{25}{9}

\frac{31}{15}

\frac{61}{16}

\frac{576}{225}

34

用前

n

项的乘积除以前

n-1

项的乘积

Divide the product of the first

n

terms by the product of the first

n-1

terms

这样便可直接把第

n

项表示成两个平方数之比

This gives the

n

th term directly as a ratio of two squares

当

n\ge2

时，第

n

项为

a_n=\frac{n^2}{(n-1)^2}\text{。}

所以

a_3=\frac{9}{4}

，且

a_5=\frac{25}{16}

，两者之和为

\frac{36}{16}+\frac{25}{16}=\frac{61}{16}

。

因此，正确答案是 C。

For

n\ge2,

the

n

th term is

a_n=\frac{n^2}{(n-1)^2}.

Therefore

a_3=\frac{9}{4}

and

a_5=\frac{25}{16},

whose sum is

\frac{36}{16}+\frac{25}{16}=\frac{61}{16}.

Therefore, the correct answer is C.

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