Question

b) Q = 11 mũ² + 13 mũ² + 15 mũ² + … + 2009 mũ².

ai nhanh nhát mình tích cho

Answer 1

Ta có: \(Q=11^2+13^2+\cdots+2009^2\)

\(=\left(1^2+2^2+\cdots+2010^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)-\left(2^2+4^2+\cdots+2010^2\right)\)

\(=\frac{2010\left(2010+1\right)\left(2\cdot2010+1\right)}{6}-\left(1+9+25+49+81\right)-2^2\left(1^2+2^2+...+1005^2\right)\)

\(=\frac{2010\cdot2011\cdot4021}{6}-165-4\cdot\frac{1005\left(1005+1\right)\left(2\cdot1005+1\right)}{6}\)

\(=335\cdot2011\cdot4021-165-4\cdot335\cdot503\cdot2011\)

\(=335\cdot2011\left(4021-4\cdot503\right)-165=335\cdot2011\cdot2009-165\)

=1353433000