Question

CM: 1/4+1/9+1/16+...+1/10000<1

Answer 1

Ta có:  \(\frac{1}{4}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{9}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{16}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

.............................................................

\(\frac{1}{10000}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)

\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< 1\)(đpcm)

Xin tk

Answer 2

Ta có: $\frac{1}{4}$+$\frac{1}{9}$+$\frac{1}{16}$+.....+$\frac{1}{10000}$=$\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$

mà $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < $\frac{1}{1.2}$+$\frac{1}{2.3}$+.......+$\frac{1}{99.100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+.......+$\frac{1}{99}$-$\frac{1}{100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1-$\frac{1}{100}$

$\Rightarrow$ $\frac{1}{2.2}$+$\frac{1}{3.3}$+.......+$\frac{1}{100.100}$ < 1

Answer 3

xl nhe em mình nó cứ vào copy phá nick của mình