Question

biết \(^{1^2+2^2+3^2+.......+50^2=42925}\)

tính nhanh tổng sau;\(A=100^2+200^2+300^2+.....+5000^2\)

Answer 1

Ta có:

\(A=100^2+200^2+300^2+...+5000^2\)

\(\Rightarrow A=\left(1.100\right)^2+\left(2.100\right)^2+\left(3.100\right)^2+...+\left(50.100\right)^2\)

\(\Rightarrow A=1^2.100^2+2^2.100^2+3^2.100^2+...+50^2.100^2\)

\(\Rightarrow A=\left(1^2+2^2+3^2+...+50^2\right).100^2\)

\(\Rightarrow A=42925.100^2\)

\(\Rightarrow A=429250000\)

Vậy A = 429250000