chứng minh rằng
\(\frac{a}{n\left(n+a\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+a}\)
áp dụng tính
A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+.....+\(\frac{1}{99.100}\)
B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}\)+....+\(\frac{5}{100.103}\)
C=\(\frac{1}{15}\)+\(\frac{1}{35}\)+.....+\(\frac{1}{2499}\)