ta có : \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-1^2\right)-\left(99^2-2^2\right)+\left(98^2-3^2\right)-...+\left(52^2-49^2\right)-\left(51^2-50^2\right)\)
\(=101\left(100-1\right)-101\left(99-2\right)+101\left(98-3\right)-...+101\left(52-49\right)-101\left(51-50\right)\)
\(=101.99-101.97+101.95-...+101.3-101.1\)
\(=101\left(99-97+95-93+...+3-1\right)\)
\(=101.\left(2+2+2+...+2\right)=101.2.25=5050\)
Mình có cách khác nha !
\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(97-97\right)+...\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\dfrac{100.101}{2}=5050\)
\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\dfrac{\left(100+1\right).100}{2}=\dfrac{101.100}{2}=\dfrac{10100}{2}=5050\)
