Question

Bài 1:

a, Tính A= \(\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right).\left(\dfrac{1}{16}-1\right).....\left(\dfrac{1}{100}-1\right).\left(\dfrac{1}{121}-1\right)\)

b, S=\(2^{2010}-2^{2009}-2^{2008}-......-2-1\)

Answer 1

a,

\(\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}........\dfrac{-99}{100}.\dfrac{-120}{121}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.........\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}\)

\(=\dfrac{1.2.3.4.....10.3.4.5.6......11.12}{2^2.3^2........11^2}\)

\(=\dfrac{1.2.11.12}{2^2.11^2}=\dfrac{12}{22}\)

Answer 2

\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\\ \Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(M=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow S=2^{2010}-M\)

* Tính M

\(M=2^{2009}+2^{2008}+...+2+1\\ \Rightarrow2^0+2^1+...+2^{2008}+2^{2009}\\ \Rightarrow2S=2^1+2^2+...+2^{2009}+2^{2010}\\ \Rightarrow2S-S=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\\ \Rightarrow S=2^{2010}-2^0=2^{2010}-1\)Thay M vào S, ta được :

\(S=2^{2010}-\left(2^{2010}-1\right)\\ \Rightarrow S=2^{2010}-2^{2010}+1\\ \Rightarrow S=1\)