Answer: B

Solution:

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ 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.