Cho B=[tex]\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}[/tex].Chứng tỏ B không phải số nguyên.
[tex]B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\\\\ =(1-\frac{1}{2^2})+(1-\frac{1}{3^2})+...+(1-\frac{1}{50^2})\\\\ =49-(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2})<49[/tex]
mặt khác: [tex]\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\\\\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\\\\ =1-\frac{1}{50}<1[/tex]
=> B> 49-1=48
=> 48<B<49 => đpcm