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
\(=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)
\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+50\right).50:2}\)
\(=\frac{1}{3.2:2}+\frac{1}{4.3:2}+...+\frac{1}{51.50:2}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)
\(=\frac{2}{1}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=\frac{2}{1}.\left(\frac{1}{2}-\frac{1}{51}\right)\)
còn lại tự tính
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)
\(=\frac{2}{2\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+...+\frac{2}{2\left(1+2+3+...+50\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2550}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(1-\frac{1}{51}\right)\)
\(=2.\frac{50}{51}\)
\(=\frac{100}{51}\)
