B=\(\frac{2^2}{1.3}\)+\(\frac{3^2}{2.4}+\frac{4^2}{3.5}+\frac{5^2}{4.6}+........+\frac{99^2}{98.100}\).TÌM PHẦN NGUYÊN CỦA B
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}\). CM:C>48
\(N=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+....+\frac{98.101}{99.100}\). CM : 97<N<98