Question

thực hiện phép tính

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

Answer 1

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+......+\dfrac{1}{1+2+......+50}\)

\(=\dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+\dfrac{1}{\dfrac{4.5}{2}}+......+\dfrac{1}{\dfrac{50.51}{2}}\)

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

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=2.\dfrac{49}{100}\)

\(=\dfrac{49}{50}\)

Answer 2

Xét thừa số tổng quát: \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)

Suy ra: \(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)

Dễ r:v