Question

Chứng minh:

\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{10^2}>0\)

Answer 1

Đặt A = 1 - 1/2² - 1/3² - 1/4² - ....... - 1/10²

=> A>1-1/2.3 - 1/3.4 - 1/4.5 - ........ - 1/10.11

=> A> 1 - (1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/10.11)

=> A> 1 - (1/2 -1/3 +1/3 - 1/4 + 1/4 -1/5+...+1/10-1/11)

=> A> 1 - (1/2 - 1/11)

=> A> 1 - 9/22

mà 9/22  < 1  nên (1 - 9/22) : dương

=> (1/9/22) > 0

=> A>0 (điều phải chứng minh)

Answer 2

\(\frac{1}{2^2}>\frac{1}{1.2};\frac{1}{3^2}>\frac{1}{2.3};.....;\frac{1}{10^2}>\frac{1}{9.10}\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-\frac{1}{1.2}-\frac{1}{2.3}-....-\frac{1}{9.10}\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-...-\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-....-\frac{1}{9}+\frac{1}{10}=\frac{1}{10}>0\)

=>ĐPCM

Answer 3

A= 1 - ( \(\frac{1}{2^2}\)+  \(\frac{1}{3^2}\)+  \(\frac{1}{4^2}\)+...+  \(\frac{1}{10^2}\))
          ________________________________
                                     B
Vì \(\frac{1}{2^2}\)<  \(\frac{1}{1.2}\);....; \(\frac{1}{10^2}\)<  \(\frac{1}{9.10}\)
=> B < \(\frac{1}{1.2}\)+  \(\frac{1}{2.3}\)+...+  \(\frac{1}{9}\)-  \(\frac{1}{10}\)
=> B < \(\frac{1}{1}\)-  \(\frac{1}{2}\)+  \(\frac{1}{2}\)-  \(\frac{1}{3}\)+...+  \(\frac{1}{9}\)-  \(\frac{1}{10}\)= \(\frac{1}{1}\)-  \(\frac{1}{10}\)= \(\frac{9}{10}\)
=> A = 1-B > 1 - \(\frac{9}{10}\)> 0
=> A > 0 (  \(Đpcm\))