Question

Chứng tỏ rằng:

D=1/2²+1/3²+1/4²+...+1/10²<1

Answer 1

Vì  giá trị  của D bé hơn 1

Answer 2

\(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(2D=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)

\(2D-D=\frac{1}{2}-\frac{1}{10^2}\)

\(D=\frac{10^2\cdot2}{10^2}-\frac{1}{10^2}=\frac{10^2\cdot2-1}{10^2}>1\)

Answer 3

sai hết vs nhau !!!!

\(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

vì \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};...;\frac{1}{10^2}< \frac{1}{9\cdot10}\)

nên \(D< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow D< 1-\frac{1}{10}\)

\(\Rightarrow D< \frac{9}{10}< 1\)

\(\Rightarrow D< 1\left(đpct\right)\)

Answer 4

M=1/1.2.3+1/2.3.4+1/3.4.5+...+1/10.11.12