Suppose that a_1=2 and the sequence (a_n) satisfies the recurrence relation (a_n-1)/(n-1)=\fraca_n-1+1(n-1)+1 for all n\ge2. What is the greatest integer less than or equal to \sum_n=1^100a_n^2?

2024 AMC 12A Problem 21 — Solution

2024 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2024 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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Concepts:recursion sum of first n squares bounding to limit cases

Difficulty rating: 2130

21.

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 $\sum_{n=1}^{100}a_n^2?$

$338{,}550$

$338{,}551$

$338{,}552$

$338{,}553$

$338{,}554$

Solution:

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ reveals $a_n=n+\tfrac1n,$ which checks out. Then $a_n^2=n^2+2+\tfrac1{n^2},$ so $\sum_{n=1}^{100}a_n^2=\sum_{n=1}^{100}n^2+200+\sum_{n=1}^{100}\frac1{n^2}=338350+200+S,$ where $1\lt S\lt2.$ Hence the sum is between $338550$ and $338552,$ and its floor is $338551.$ Thus, the correct answer is B.

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