Question

1^{2 +} 3^{2 +} 5^{2 + ... +} 99² 

^{giúp em với ạ}

Answer 1

Ta có: \(1^2+3^2+5^2+\cdots+99^2\)

\(=1^2+2^2+\cdots+100^2-\left(2^2+4^2+\cdots+100^2\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)-2^2\left(1^2+2^2+\cdots+50^2\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}-4\cdot\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}\)

\(=\frac{100\cdot101\cdot201}{6}-\frac{2\cdot50\cdot51\cdot101}{3}=50\cdot101\cdot67-100\cdot17\cdot101\)

\(=101\cdot50\left(67-2\cdot17\right)=5050\cdot\left(67-34\right)=33\cdot5050\)

=166650