a. Ta có: \(A=1\cdot3+3\cdot5+5\cdot7+...+99\cdot101\)
\(\Rightarrow A=1\left(1+2\right)+3\cdot\left(3+2\right)+...+99\left(99+2\right)\)
\(\Rightarrow A=\left(1^2+3^2+5^2+...+97^2+99^2\right)+2\left(1+3+5+...+97+99\right)\)
Đặt \(M=1^2+3^2+5^2+99^2\)
\(\Rightarrow M=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+50^2\right)\)
Tính dãy tổng quát \(N=1^2+2^2+3^2+...+n^2\)
\(\Rightarrow N=1\left(0+1\right)+2\left(1+1\right)+3\left(2+1\right)+...+n[\left(n-1\right)+1]\)
\(\Rightarrow N=\left[1\cdot2+2\cdot3+...+\left(n-1\right)n\right]+\left(1+2+3+...+n\right)\)
\(\Rightarrow N=n\left(n+1\right)\cdot\left[\left(n-1\right):3+1:2\right]=n\left(n+1\right)\cdot\left(2n+1\right):6\)
Áp dụng vào M ta được:
\(M=100\cdot101\cdot201:6-4\cdot50\cdot51\cdot101:6=166650\)
\(\Rightarrow A=166650+2\left(1+99\right)\cdot50:2\)
\(\Rightarrow A=166650+5000=171650\)
Vậy \(A=171650\)
