Find \((1^2 + 1 \cdot 2 + 2^2) + (2^2 + 2 \cdot 3 + 3^2) + \cdots + (99^2 + 99 \cdot 100 + 100^2).\)
Thanks!
+486 If you use the square of a binomial, you get $(1^2+1*2+2^2)+(2^2+2*3+3^2)+...+(99^2+99*100+100^2)=(1+2)^2+(2+3)^2+(3+4)^2+...+(99+100)^2=3^2+5^2+7^2+...+197^2+199^2$. Now, this is equivalent to the sum of the first odd squares, minus $1^2$. The formula for the sum of the first n odd perfect squares is $\frac{n(2n+1)(2n-1)}{3}$. Now, let's say we include the $1^2$ so we can use the formula, just remember to subtract $1^2$ at the end. This is then the first $100$ perfect squares. Using the formula, we get $\frac{100(201)(199)}{3}=\frac{3999900}{3}=1333300$. But wait! We need to subtract $1^2$, because we added it to use the formula. So our final answer is $1333300-1^2=\boxed{1333299}$
