Question

[David Santas]

Tính: A = (1- 1/2)(1-1/3)(1-1/4)...(1-1/2016)(1-1/2017)

S= 2^2010 - 2^2009 - 2^2008 - ... - 2 - 1

Answer 1

[Diệu Huyền]

Câu 1:

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\left(1-\frac{1}{2017}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}.\frac{2016}{2017}\)

\(A=\frac{1}{2017}\)

Vậy ..............................

Phần giống nhau là gạch ý!

Câu 2

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

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

Đặt \(Q=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2Q=2^{2010}+2^{2009}+...+4+2\)

\(\Rightarrow2Q-Q=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(\Rightarrow Q=2^{2010}-1\)

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

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

\(\Rightarrow S=1\)

Vậy .........................

Answer 2

[Vũ Minh Tuấn]

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

\(\Rightarrow2S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)

\(\Rightarrow2S-S=\left(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\right)-\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)

\(\Rightarrow S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2-2^{2010}+2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow S=2^{2011}-2^{2010}-2^{2010}+1\)

\(\Rightarrow S=2^{2011}-2.2^{2010}+1\)

\(\Rightarrow S=2^{2011}-2^{2011}+1\)

\(\Rightarrow S=0+1\)

\(\Rightarrow S=1.\)

Vậy \(S=1.\)

Chúc bạn học tốt!