Question

Tính giá trị biểu thức: A=1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) +...+ 1/(1+2+3+4+5+...+49+50) =?

Tĩnh giùm nha

Answer 1

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)

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

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)

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

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)