Ta có:
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...\frac{1}{50.51}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...-\frac{1}{50}+\frac{1}{50}-\frac{1}{51}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{51}\)
Vì \(\frac{1}{2}-\frac{1}{51}<1\)
nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<1\)
\(y<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)
\(y<1-\frac{49}{50}<1\)
=> y < 1
cac cau gui hinh nhe! to khong doc duoc :(
Biểu thức trên đâu bé hơn \(\frac{1}{2}\) -\(\frac{1}{51}\)
