Question

[Trương Quỳnh Trang]

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

(Các bạn giải ra các bước hộ mình nhé. Cám ơn các bạn nhiều)

Answer 1

[Hồ Thu Giang]

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)

=> x + 1 = 1993

=> x = 1993 - 1

=> x = 1992