1. The sum of the squares of 50 consecutive odd integers is 300850 . Find the largest odd integer whose square is the last term of this sum.
1. Answer: 121
Solution. Let the integers be $X+2 X+4 \cdots X+100$. Then
$$
(X+2)^{2}+(X+4)^{2}+\cdots+(X+100)^{2}=300850
$$

Let $y=X+51$ and regrouping the terms, we obtain
$$
\left[(y-49)^{2}+(y+49)^{2}\right]+\left[(y-47)^{2}+(y+47)^{2}\right]+\cdots+\left[(y-1)^{2}+(y+1)^{2}\right]=300850
$$
which simplifies to
$$
50 y^{2}+2\left(1^{2}+3^{2}+5^{2}+7^{2}+\cdots+49^{2}\right)=300850
$$

Using the fact that $1^{2}+3^{2}+5^{2}+\cdots+(2 n-1)^{2}=\frac{4}{3} n^{3}-\frac{1}{3} n$, we obtain $y=72$. Hence $X=21$, so that the required number is 121 .