Question

3/2+3/6+3\12+3/20+...+3/9900

Answer 1

nhanh nhanh đang cần

Answer 2

Đặt \(A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\)

\(=\frac{3}{1\times2}+\frac{3}{2\times3}+\frac{3}{3\times4}+...+\frac{3}{99\times100}\)

\(\Rightarrow A:3=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}\times3=\frac{297}{100}\)

Vậy \(A=\frac{297}{100}\).

Answer 3

mình chịu lun .xin lỗi!

Kocho Shinobu