Question

[phùng thị khánh huyền]

Cmr P=1/2^2 +1/3^2 +1/4^4 + ... 1/100^2  <1

Answer 1

[Nguyễn Phương Uyên]

\(P=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow P< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow P< 1-\frac{1}{100}\)

\(\Rightarrow P< \frac{99}{100}< 1\)

Answer 2

[Huy Hoang]

\(P=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}+\frac{1}{100}\)

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

Vậy : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< 1\left(đpcm\right)\)