Ta thấy đc quy luật:
\(\frac{2^2-1^2}{2^2}=\frac{2+1}{2+2}=\frac{3}{4}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}=\frac{6+2}{6+3}=\frac{8}{9}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}=\frac{12+3}{12+4}=\frac{15}{16}\)
Nên:
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}+...+\frac{100^2-99^2}{9900^2}=\frac{9900+99}{9900+100}=\frac{9999}{10000}\)
Hay A<1(đpcm)