Question

[nghĩa phan]

3/1 +  3/+1+2 + 3/1+2+3 + 3+1+2+3+4 + ... + 3/1+2+3+...+100

Answer 1

[Pencil]

Ta đặt biểu thức trên là: A

\(\dfrac{1}{3}A=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+100}\)

\(\dfrac{1}{3}A=1+\dfrac{1}{\dfrac{2\left(1+2\right)}{2}}+\dfrac{1}{\dfrac{3\left(1+3\right)}{2}}+...+\dfrac{1}{\dfrac{100\left(1+100\right)}{2}}\)

\(\dfrac{1}{3}A=1+\dfrac{2}{2\left(1+2\right)}+\dfrac{2}{3\left(1+3\right)}+...+\dfrac{2}{100\left(1+100\right)}\)

\(\dfrac{1}{3}A=1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{100.101}\)

\(\dfrac{1}{6}A=\dfrac{1}{2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{101-100}{100.101}\)

\(\dfrac{1}{6}A=\dfrac{1}{2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{101}{100.101}-\dfrac{100}{100.101}\)

\(\dfrac{1}{6}A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\dfrac{1}{6}A=1-\dfrac{1}{101}\)

\(\dfrac{1}{6}A=\dfrac{100}{101}\)

\(A=\dfrac{100}{101}\div\dfrac{1}{6}\)

\(A=\dfrac{600}{101}\)