Question

Answer 1

Lời giải:

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

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

.....

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

Do đó:\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)

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

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

Ta có đpcm.

Answer 2

Lời giải:

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

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

.....

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

Do đó:\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)

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

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

Ta có đpcm.