Question

[Sakura Kinomoto]

Tính tổng:

a) A=\(\left(1-\frac{1}{2}\right)\)\(\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{3}\right)\)........\(\left(1-\frac{1}{100}\right)\)

b) B=\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ .... +\(\frac{1}{2014.2015}\)

c) C=\(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ .... +\(\frac{2}{97.99}\)

d. D=\(\frac{2}{1.4}\)+ \(\frac{2}{4.7}\)+ ..... +\(\frac{2}{2012.2015}\)

Answer 1

[Mạnh Lê]

b) \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}\)

\(B=1-\frac{1}{2015}\)

\(B=\frac{2014}{2015}\)

Answer 2

[Duong Minh Hieu]

a) \(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(=\frac{1}{100}\)

b)\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)

còn lại tự giải nha gần giống như phần b thôi cũng thú vị.

ủng hộ nha