Ta có: \(C=1\cdot3^2+3\cdot5^2+\cdots+97\cdot99^2\)
\(=3^2\left(3-2\right)+5^2\left(5-2\right)+\cdots+99^2\left(99-2\right)\)
\(=\left(3^3+5^3+\cdots+99^3\right)-2\left(3^2+5^2+\cdots+99^2\right)\)
Đặt \(A=3^3+5^3+...+99^3\)
\(=\left(1^3+3^3+\cdots+99^3\right)-1^3\)
\(=\left(1^3+2^3+\cdots+99^3+100^3\right)-\left(2^3+4^3+\cdots+100^3\right)-1^3\)
\(=\left(1+2+\cdots+100\right)^2-2^3\left(1^3+2^3+\cdots+50^3\right)-1\)
\(=\left(100\cdot\frac{101}{2}\right)^2-8\cdot\left(1+2+3+\cdots+50\right)^2-1\)
\(=\left(50\cdot101\right)^2-8\left(50\cdot\frac{51}{2}\right)^2-1=5050^2-\left(25\cdot51\right)^2\cdot8-1\)
\(=5050^2-8\cdot1275^2-1=12497499\)
Đặt \(B=3^2+5^2+\cdots+99^2\)
\(=\left(1^2+3^2+5^2+\cdots+99^2\right)-1\)
\(=\left(1^2+2^2+\cdots+100^2\right)-\left(2^2+4^2+\cdots+100^2\right)-1\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}-2^2\left(1^2+2^2+\cdots+50^2\right)-1\)
\(=\frac{100\cdot101\cdot201}{6}-4\cdot\frac{50\left(50+1\right)\left(2\cdot50+1\right)}{6}-1\)
\(=\frac{100\cdot101\cdot201}{6}-\frac{4\cdot50\cdot51\cdot101}{6}-1=\frac{100\cdot101\cdot201-200\cdot51\cdot101}{6}-1\)
\(=\frac{100\cdot101\left(201-51\cdot2\right)}{6}-1=\frac{100\cdot101\cdot99}{6}-1\)
\(=50\cdot101\cdot33-1=166649\)
C=A-2B
=12497499-2*166649
=12164201
