Question

Tính S=\(\left(1-\dfrac{1}{1+2}\right).\left(1-\dfrac{1}{1+2+3}\right).\left(1-\dfrac{1}{1+2+3+4}\right).....\left(1-\dfrac{1}{1+2+...+2014}\right)\)

Answer 1

Giải:

\(S=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+...+2014}\right)\)

Nhận xét:

\(1-\dfrac{1}{1+2+3+...+n}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n+1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Do đó:

\(S=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{49.52}{50.51}\)

\(\Rightarrow S=\dfrac{\left(1.2.3....49\right).\left(4.5.6....52\right)}{\left(2.3.4....50\right).\left(3.4.5....51\right)}\)

\(\Rightarrow S=\dfrac{1.4.52}{50.3}=\dfrac{104}{75}\)