Question

Chứng tỏ rằng :

\(A=\dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + ...+ \dfrac{1}{100} < 1\)

Answer 1

\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)

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

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

\(A< \frac{9}{10}\Rightarrow A< 1\left(đpcm\right)\)

Answer 2

Viết hơi rắc rối,  ko hiểu=ib.

Ta có:

A=1/4+1/9+1/16+...+1/100

=>A=1/2²+1/3²+1/4²+...+1/10²

=>A<1/(1.2)+1/(2.3)+1/(3.4)+...+1/(9.10)         =1-1/2+1/2-1/3+...+1/9-1/10

         =1-1/10=9/10<1

=>A<1(đpcm)