Question

a. Chứng minh rằng

\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\in Nsao\right)\)

b. Áp dụng câu a tính:

A= \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

B= \(\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)

C= \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

Answer 1

b) A=1/2.3+1/3.4+....+1/99.100

=> A=1/2-1/3+1/3-1/4+....+1/99-1/100

=> A=1/2-1/100

=> A=50/100-1/100

=> A=49/100

Answer 2

49/100 

k nhe

Answer 3

49/100

Answer 4

a) Xét vế phải \(\frac{1}{n}\) - \(\frac{1}{n+a}\) = \(\frac{n+a}{n\left(n+a\right)}\) - \(\frac{n}{n\left(n+a\right)}\) = \(\frac{n+a-n}{n\left(n+a\right)}\) = \(\frac{a}{n\left(n+a\right)}\) => đpcm

b) \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) +...+\(\frac{1}{99.100}\)

= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) +...+\(\frac{1}{99}\) - \(\frac{1}{100}\)

= \(\frac{1}{2}\) - \(\frac{1}{100}\) = \(\frac{49}{100}\)

B=\(\frac{5}{1.4}\) + \(\frac{5}{4.7}\) +...+\(\frac{5}{100.103}\)

= \(\frac{5}{3}\) . (1- \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{7}\) +...+ \(\frac{1}{100}\) - \(\frac{1}{103}\)) = \(\frac{5}{3}\) . ( 1-\(\frac{1}{103}\)) = \(\frac{5}{3}\) . \(\frac{102}{103}\) = \(\frac{170}{103}\) 

C= \(\frac{1}{15}\) + \(\frac{1}{35}\) +...+ \(\frac{1}{2499}\) = \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) +...+\(\frac{1}{49.51}\) = \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) +...+ \(\frac{1}{49}\) - \(\frac{1}{50}\) = \(\frac{1}{3}\) - \(\frac{1}{50}\) = \(\frac{47}{150}\)