Question

设 $a_1=2$ ，且序列 $(a_n)$ 对所有 $n\ge2$ 满足递推关系 $\frac{a_n-1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}\text{。}$ 求不超过下式的最大整数： $\sum_{n=1}^{100}a_n^2\text{？}$

Suppose that $a_1=2$ and the sequence $(a_n)$ satisfies the recurrence relation $\frac{a_n-1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}$ for all $n\ge2.$ What is the greatest integer less than or equal to $\sum_{n=1}^{100}a_n^2?$

$338{,}550$

$338{,}551$

$338{,}552$

$338{,}553$

$338{,}554$

答案：B

知识点：递推前n项平方和极限情形界定

难度评级：2130

小提示：

计算前几项并猜测通项 $a_n=n+\dfrac1n$ 。

Compute a few terms and conjecture the closed form $a_n=n+\dfrac1n$

大提示：

于是 $a_n^2=n^2+2+\dfrac1{n^2}$ ，且 $\displaystyle\sum_{n=1}^{100}\dfrac1{n^2}$ 严格介于 $1$ 和 $2$ 之间

Then $a_n^2=n^2+2+\dfrac1{n^2},$ and $\displaystyle\sum_{n=1}^{100}\dfrac1{n^2}$ lies strictly between $1$ and $2$

解答：

递推式可化为 $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1)$ 。计算前几项 $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ ，可以猜出 $a_n=n+\tfrac1n$ 。这可以用归纳法证明：把 $a_{n-1}=n-1+\tfrac1{n-1}$ 代入递推式得 $a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})$ $=n+\tfrac1n$ 。于是 $a_n^2=n^2+2+\tfrac1{n^2}$ ，所以 $\begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S \end{aligned}\text{，}$ 其中 $S\gt1$ 且 $S\lt1+\int_1^\infty x^{-2}\,dx=2$ 。因此该和介于 $338551$ 与 $338552$ 之间，其向下取整为 $338551$ 。因此正确答案是 B。

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ suggests $a_n=n+\tfrac1n.$ This follows by induction: substituting $a_{n-1}=n-1+\tfrac1{n-1}$ into the recurrence gives $a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})$ $=n+\tfrac1n.$ Then $a_n^2=n^2+2+\tfrac1{n^2},$ so $\begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S, \end{aligned}$ where $S\gt1$ and $S\lt1+\int_1^\infty x^{-2}\,dx=2.$ Hence the sum is between $338551$ and $338552,$ and its floor is $338551.$ Thus, the correct answer is B.

2024 AMC 12A 第 21 题

答案：B

难度评级：2130

小提示：

大提示：

解答：

其他年份的第 21 题