Find $1^2+3^2+...99^2$ given $1^1+2^2+...+100^2$ and $1^1+2^2+...+50^2$ We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?

This should help: $$2^2+4^2+6^2+\dots+100^2=2^2(1^2+2^2+3^2+\dots+50^2)=\dots$$
