Question

[Hoang phuong]

Tính A= 1/3 + 1/(3+6)+1/(3+6+9)+....+ 1/(3+6+9+...+2013)

Answer 1

[soyeon_Tiểu bàng giải]

\(A=\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2013}\)

\(A=\frac{1}{3}.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...671}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+671\right).671:2}\right)\)

\(A=\frac{1}{3}.\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{671.672}\right)\)

\(A=\frac{1}{3}.2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{671.672}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{671}-\frac{1}{672}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{672}\right)\)

\(A=\frac{2}{3}.\frac{671}{672}=\frac{671}{1008}\)