Question

chứng minh rằng 1 phần 2² + 1 phần 3² + 1 phần 4² + ... + 1 phần 100² <1

Answer 1

\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}\)

....

\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)

do đó \(A<\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)

=>A<1

Answer 2

sẽ là 1/4+1/9+1/16........tổng sẽ ko lớn hơn 1

Answer 3

???????????????????????

Answer 4

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2}\);\(\frac{1}{3^2}<\frac{1}{2.3}\);............;\(\frac{1}{100^2}\)\(<\frac{1}{99.100}\)

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

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{99.100}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)\(=\frac{99}{100}<1\)

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