Question

Tính nhanh

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+50}\)

Trần Thùy Dung đâu ! giúp

Answer 1

11+2 +11+2+3 +...+11+2+3+...+50 

=22(1+2) +22(1+2+3) +...+22(1+2+3+...+50) 

=26 +212 +220 +...+22550 

=2(11.2 +12.3 +13.4 +...+150.51 )

=2(1−12 +12 −13 +13 −14 +...+150 −151 )

=2(1−151 )

=2.5051 

=10051 

Answer 2

ai làm giúp mik cảm ơn nhiều

Answer 3

\(=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)

\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+50\right).50:2}\)

\(=\frac{1}{3.2:2}+\frac{1}{4.3:2}+...+\frac{1}{51.50:2}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)

\(=\frac{2}{1}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=\frac{2}{1}.\left(\frac{1}{2}-\frac{1}{51}\right)\)

còn lại tự tính

Answer 4

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)

\(=\frac{2}{2\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+...+\frac{2}{2\left(1+2+3+...+50\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2550}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(1-\frac{1}{51}\right)\)

\(=2.\frac{50}{51}\)

\(=\frac{100}{51}\)