\(1^2+2^2+3^2+...+n^2\\ =1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+n\left(n+1-1\right)\\ =1\cdot 2+2\cdot 3+3\cdot 4+...+n\left(n+1\right)-1-2-3-...-n\\ =\frac{3\left[1\cdot 2+2\cdot 3+3\cdot 4+...+n\left(n+1\right)\right]}{3}-\left(1+2+3+...+n\right)\\ =\frac{1\cdot 2\cdot \left(3-0\right)+2\cdot 3\cdot \left(4-1\right)+3\cdot 4\cdot \left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]}{3}-\frac{n\left(n+1\right)}{2}\\ =\frac{1\cdot 2\cdot 3-0\cdot 1\cdot 2+2\cdot 3\cdot 4-1\cdot 2\cdot 3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)}{3}-\frac{n\left(n+1\right)}{2}\\ =\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\\ =\frac{2n\left(n+1\right)\left(n+2\right)-3n\left(n+1\right)}{6}\\ =\frac{n\left(n+1\right)\left[2\left(n+2\right)-3\right]}{6}\\ =\frac{n\left(n+1\right)\left(2n+4-3\right)}{6}\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)\(A=2^2+4^2+6^2+...+100^2\\ =2^2\cdot1^2+2^2\cdot2^2+2^2\cdot3^2+...+2^2\cdot50^2\\ =2^2\left(1^2+2^2+3^2+...+50^2\right)\\ =4\cdot\dfrac{50\cdot51\cdot101}{6}\\ =4\cdot42925\\ =171700\)
\(A=2^2+4^2+6^2+.......+98^2+100^2\)
\(\Leftrightarrow\dfrac{1}{4}A=1^2+2^2+3^2+.......+49^2+50^2\)
\(\Leftrightarrow\dfrac{1}{4}A=\dfrac{50.51\left(2.50+1\right)}{6}\)
\(\Leftrightarrow\dfrac{1}{4}A=42925\)
\(\Leftrightarrow A=171700\)
