Question

A=\(\frac{1}{30.31}+\frac{1}{31.32}+\cdots+\frac{1}{n\left(n+1\right)}\)

Answer 1

`A = 1/(30.31) + 1/(31.32) + ... + 1/(n(n+1))`

`A = 1/30 - 1/31 + 1/31 - 1/32 + ... + 1/n - 1/(n+1) `

`A = 1/30 - 1/(n+1)`

`A = (n - 29)/(30(n+1))`

Answer 2

A=\(\frac{\mathbb{N}}{\mathbb{N}+1}\)

Answer 3

\(A=\frac{1}{30.31}+\frac{1}{31.32}+\cdots+\frac{1}{n\left(n+1\right)}\)

\(A=\frac{1}{30}-\frac{1}{31}+\frac{1}{31}-\frac{1}{32}+\cdots+\frac{1}{n}-\frac{1}{n+1}\)

\(A=\frac{1}{30}-\frac{1}{n+1}=\frac{n+1}{30\left(n+1\right)}-\frac{30}{n+1}\)

\(A=\frac{\left(n+1\right)-30}{30\left(n+1\right)}=\frac{n-29}{30n+30}\)

Vậy \(A=\frac{n-29}{30n+30}\)