2024 AMC 12A Problem 21Suppose that a1=2a_1=2a1=2 and the sequence (an)(a_n)(an) satisfies the recurrence relation an−1n−1=an−1+1(n−1)+1\frac{a_n-1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}n−1an−1=(n−1)+1an−1+1 for all n≥2n\ge2n≥2. What is the greatest integer less than or equal to ∑n=1100an2 ?\displaystyle\sum_{n=1}^{100} a_n^2\ ?n=1∑100an2 ?(A) 338,550\text{(A)}\;338,550(A)338,550(B) 338,551\text{(B)}\;338,551(B)338,551(C) 338,552\text{(C)}\;338,552(C)338,552(D) 338,553\text{(D)}\;338,553(D)338,553(E) 338,554\text{(E)}\;338,554(E)338,554Related TopicsCoreTelescoping Approaches to Solving RecurrencesTelescopingMajorSums of PowersMinorArithmetic Sequences and SeriesCheck AnswerYour answer:ABCDECheckHints (8)Hint 1nan−n=(n−1)an−1+(n−1)na_n-n=(n-1)a_{n-1}+(n-1)nan−n=(n−1)an−1+(n−1)Hint 2Let nan=bnna_n=b_nnan=bn, so b1=a1=2b_1=a_1=2b1=a1=2.Hint 3By Hints 1 and 2bn−bn−1=2n−1b_n-b_{n-1}=2n-1bn−bn−1=2n−1Hint 4⇒bn−b1=(3+2n−1)(n−1)2=n2−1\Rightarrow\quad b_n-b_1=\frac{(3+2n-1)(n-1)}{2}=n^2-1⇒bn−b1=2(3+2n−1)(n−1)=n2−1bn=n2+1b_n=n^2+1bn=n2+1Hint 5bn=nan⇒an=bnn=n2+1n=n+1nb_n=na_n\quad\Rightarrow\quad a_n=\frac{b_n}{n}=\frac{n^2+1}{n}=n+\frac{1}{n}bn=nan⇒an=nbn=nn2+1=n+n1Hint 6A=∑n=1100an2=∑n=1100(n+1n)2A=\sum_{n=1}^{100}a_n^2=\sum_{n=1}^{100}\left(n+\frac{1}{n}\right)^2A=n=1∑100an2=n=1∑100(n+n1)2=∑n=1100(n2+2+1n2)=\sum_{n=1}^{100}\left(n^2+2+\frac{1}{n^2}\right)=n=1∑100(n2+2+n21)Hint 7C=∑n=21001n2<∑n=21001n(n−1)=∑n=2100(1n−1−1n)=BC=\displaystyle\sum_{n=2}^{100}\frac{1}{n^2}<\displaystyle\sum_{n=2}^{100}\frac{1}{n(n-1)}=\displaystyle\sum_{n=2}^{100}\left(\frac{1}{n-1}-\frac{1}{n}\right)=BC=n=2∑100n21<n=2∑100n(n−1)1=n=2∑100(n−11−n1)=BSo,B=1−1100=99100B=1-\frac{1}{100}=\frac{99}{100}B=1−1001=10099Hint 8A=∑n=1100(n2+2+1n2)A=\displaystyle\sum_{n=1}^{100}\left(n^2+2+\frac{1}{n^2}\right)A=n=1∑100(n2+2+n21)=(12+22+32+⋯+1002)+200+112+C=(1^2+2^2+3^2+\cdots+100^2)+200+\frac{1}{1^2}+C=(12+22+32+⋯+1002)+200+121+C<100(101)(201)6+201+B<\frac{100(101)(201)}{6}+201+B<6100(101)(201)+201+B=338551+99100=338551+\frac{99}{100}=338551+10099Ans=338551\text{Ans}=338551Ans=338551Related Problems (7)AMC 10B 2022 (Problem 9)AMC 8 2022 (Problem 8)AMC 12B 2025 (Problem 7)MATHCOUNTS 2026 State Sprint Round (Problem 24)AIME II 2025 (Problem 4)AMC 12A 2023 (Problem 20)AMC 10B 2025 (Problem 18)Final Answer(B) 338,551
