AMC12 2024 A

Multiply both sides of the recurrence to find that $n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)$. Let $b_n=n(a_n-1)$. Then the previous relation becomes \[b_n=b_{n-1}+2(n-1)\] We can rewrite this relation for values of $n$ until $1$ and use telescoping to derive an explicit formula: \[b_n=b_{n-1}+2(n-1)\] \[b_{n-1}=b_{n-2}+2(n-2)\] \[b_{n-2}=b_{n-3}+2(n-3)\] \[\cdot\] \[\cdot\] \[\cdot\] \[b_2=b_1+2(1)\] Summing the equations yields: \[b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)\] \[b_n-b_1=2\cdot\frac{n(n-1)}{2}\] \[b_n-1=n(n-1)\] \[b_n=n(n-1)+1\] Now we can substitute $a_n$ back into our equation: \[n(a_n-1)=n(n-1)+1\] \[a_n-1=n-1+\frac{1}{n}\] \[a_n=n+\frac{1}{n}\] \[a_n^2=n^2+\frac{1}{n^2}+2\] Thus the sum becomes \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2\] We know that $\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350$, and we also know that $\sum^{100}_{n=1} 2=200$, so the requested sum is equivalent to $\sum^{100}_{n=1} \frac{1}{n^2}+338550$. All that remains is to calculate $\sum^{100}_{n=1} \frac{1}{n^2}$, and we know that this value lies between $1$ and $2$ (see the note below for a proof). Thus, \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552\] \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551\] so \[\sum^{100}_{n=1} a_n^2\in(338551,338552)\] and thus the answer is $\boxed{\textbf{(B) }338551}$. Note: $\sum^{100}_{n=1} \frac{1}{n^2}< \sum^{\infty}_{n=1} \frac{1}{n^2}=\frac{\pi^2}{6}<2$. It is obvious that the sum is greater than 1 (since it contains $\frac{1}{1^2}$ as one of its terms).