Question

Tính:

a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

b) \(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)

Answer 1

a) \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)

\(A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=\frac{1}{1.2}-\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{9900}\)

\(A=\frac{9898}{19800}.\)

Vậy :

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(A=\frac{9898}{19800}:2\)

\(A=\frac{4949}{19800}.\)

 

Answer 2

a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

A = \(\frac{1}{2}.\frac{4949}{9900}\)

A = \(\frac{4949}{19800}\)

Answer 3

tính dc A chứ B chưa hc

Answer 4

Dạng toán lớp 5 

Answer 5

mk đúng oy dạng B mk chưa hc