Question

So sánh S =\(\frac{2}{1×2×3}+\frac{2}{2×3×4}+\frac{2}{3×4×5}+...+\frac{2}{2010×2011×2012}\) với P=\(\frac{1}{2}\)

Answer 1

S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2010.2011.2012}\)

  =\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)

  =\(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)(Vì \(\frac{1}{2011.2012}>0\))

=> S <\(\frac{1}{2}\)

Answer 2

\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{2010.2011.2012}\)

\(S=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2012-2010}{2010.2011.2012}\)

\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)

\(S=\frac{1}{1.2}-\frac{1}{2011.2012}=\frac{2023065}{4046132}\)

\(\text{Vì}\)\(\frac{2023065}{4046132}< \frac{1}{2}\Rightarrow S< P\)