Question

[Cậu Bé Ngu Ngơ]

Cho E=(1-1/1+2)(1-1/1+2+3)(1-1/1+2+3)...(1-1/1+2+3+...+n)

và F=n+2/n với n thuộc N^* .Tính E/F

Answer 1

[soyeon_Tiểubàng giải]

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

\(E=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}.\frac{9}{\left(1+4\right).4:2}...\frac{\left(1+n\right).n:2-1}{\left(1+n\right).n:2}\)

\(E=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{2.\left[\left(1+n\right).n:2-1\right]}{n.\left(n+1\right)}\)

\(E=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)

\(E=\frac{1.2.3...\left(n-1\right)}{2.3.4...n}.\frac{4.5.6...\left(n+2\right)}{3.4.5...\left(n+1\right)}\)

\(E=\frac{1}{n}.\frac{n+2}{3}=\frac{n+2}{3n}\)

\(\frac{E}{F}=\frac{n+2}{3n}:\frac{n+2}{n}=\frac{n+2}{3n}.\frac{n}{n+2}=\frac{1}{3}\)

Answer 2

[Cậu Bé Ngu Ngơ]

Ai đúng tôi tick cho