Question

[I love G Friend]

1. Hãy tính tổng:

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

2. Chứng minh:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\)

Answer 1

[Nguyễn Hữu Huy]

1 : dễ mà 

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

1 phần 1 - 1 phần 2 = 1 phần 1.2 mà tương tự như thế đó

=> 1 - 1 phần n+1 

đS

Answer 2

[Nguyễn Hưng Phát]

\(\frac{1}{1.2}+\frac{1}{2.3}+..........+\frac{1}{n.\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(=\frac{n}{n+1}\)

Bài 2:Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};.................;\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+.........+\frac{1}{\left(n-1\right).n}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{n-1}-\frac{1}{n}\)

=\(1-\frac{1}{n}<1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{n^2}<1\)