Question

[xin cảm ơn]

tính 1^2+3^2+5^2+...+97^2+99^2

Answer 1

[Nguyễn Linh Chi]

Mình có công thức: 

\(1^2+2^2+3^2+...+n^2\)

\(=1^2\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+n\left(n+1-1\right)\)

\(=\left(1.2+2.3+3.4+...+n\left(n+1\right)\right)-\left(1+2+3+...+n\right)\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{\left(n+1\right).n}{2}\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

=> \(1^2+2^2+3^2+4^2+...+98^2+99^2=\frac{99.100.199}{6}\)

và \(1^2+2^2+3^2+...+49^2=\frac{49.50.99}{6}\)

Khi đó: 

\(1^2+3^2+5^2+...+97^2+99^2\)

\(=\left(1^2+2^2+3^2+4^2+5^2+...+97^2+98^2+99^2\right)\)

\(-\left(2^2+4^2+6^2+...+98^2\right)\)

\(=\left(1^2+2^2+3^2+4^2+5^2+...+97^2+98^2+99^2\right)\)

\(-4\left(1^2+2^2+3^2+...+49^2\right)\)

\(=\frac{99.100.199}{6}-4.\frac{49.50.99}{6}=166650\)