Question

[Hoang Phươngpsh]

tính nhanh

D= 1/2 + 1/2*3 + 1/3*4 + ... + 1/99*100

Answer 1

[thien ty tfboys]

\(D=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{99.100}\)

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

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

\(D=\frac{99}{100}\)

Vậy tổng D bằng \(\frac{99}{100}\)

Answer 2

[Nguyễn Thị BÍch Hậu]

tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

áp dụng ta có: \(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)