Question

chứng minh : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)  bé hơn 1

Answer 1

Nhận xét:

\(\frac{1}{2^2}<\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3^2}<\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)

....

\(\frac{1}{10^2}<\frac{1}{10\times11}=\frac{1}{10}-\frac{1}{11}\)

Tính tổng ta có:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}<1\)

Answer 2

đặt A=1/1.2+1/2.3+...+1/9.10

B=1/2^2+1/3^2+...+1/10^2

ta có:B=1/2^2+1/3^2+...+1/10^2<A=1/1.2+1/2.3+...+1/9.10

mà A=1/1.2+1/2.3+...+1/9.10

=1-1/2+1/2-1/3+...+1/9-1/10

=1-1/10<1

=>A<B<1

=>A<1

Answer 3

C2:\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{10^2}\)<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

=\(1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}<1\)

=\(\frac{9}{10}<1=\frac{9}{10}<\frac{10}{10}\)

Vậy \(\frac{9}{10}<1\)

Answer 4

C2:Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}<\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)

=\(1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)<1

=9/10<1=9/10<10/10

Vậy 9/10<1

Answer 5

1/2^2+1/3^2+1/4^2+....+1/10^2<1

Đặt là A

A<B=1/1.2+1/2.3+1/3.4+...+1/9.10

     B=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

     B=1-1/10=910<1

 Do A< B<1

 Suy ra A<1

Answer 6

TA có:1/2^2+1/3^2+1/4^2+...+1/10^2<1/1.2+1/2.3+1/3.4+...+1/9.10=1-1/10=9/10

Do 9/10<1 suy ra 1/2^2+1/3^2+...+1/10^2