Question

Tính:

4+12+24+40+...+19404+19800=?

Answer 1

Đặt \(A=4+12+.....+19800\)

\(\frac{1}{2}A=2+6+........+9900\)

\(\frac{1}{2}A=1.2+2.3+.........+99.100\)

\(\frac{3}{2}A=1.2.3+2.3.\left(4-1\right)+......+99.100.\left(101-98\right)\)

\(\frac{3}{2}A=1.2.3+2.3.4-1.2.3+.........+99.100.101-98.99.100\)

\(\frac{3}{2}A=99.100.101\)

\(\Rightarrow A=99.100.101:\frac{3}{2}\)

\(\Rightarrow A=666600\)