Question

Chứng tỏ rằng:

a)  \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<2\)

b)  \(B=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{39}+\frac{1}{40}.\) Chứng tỏ \(\frac{1}{2}\)< B < 1

c)  \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}<\frac{1}{100}\)

Answer 1

a)đặt B=1/2.3+1/3.4+...+1/99.100

=1/1.2+1/2.3+1/3.4+...+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100<1 (1)

Mà 1<2(2)

A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)

từ (1),(2),(3) =>A<2

b,c tự làm

Answer 2

Thế mà ko biết làm

Answer 3

Thế mà ko biết làm

Answer 4

a) \(\frac{1}{2.2}<\frac{1}{1.2}\)

    \(\frac{1}{3.3}<\frac{1}{2.3}\)

....

    \(\frac{1}{100.100}<\frac{1}{99.100}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

->\(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=> A < 2- \(\frac{1}{100}\)

-> A < 2- \(\frac{1}{100}<2\)

--> A <2 

Answer 5

\(\frac{1}{20}>\frac{1}{21}\)

\(\frac{1}{20}>\frac{1}{22}\)

...

\(\frac{1}{20}>\frac{1}{40}\)

==> \(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}>\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\)

---> \(\frac{20}{20}>B\)

-->1 >B

ta có 

\(\frac{1}{21}<\frac{1}{40}\)

\(\frac{1}{22}<\frac{1}{40}\)

....

\(\frac{1}{40}=\frac{1}{40}\)

--> \(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}<\frac{20}{40}\)

--> B \(<\frac{1}{2}\)

Vay 1/2 < B < 1