Question

Tính giá trị của biểu thức:

\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)

\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)

Answer 1

a) C=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)

\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}\)\(=\frac{1.2.3...999}{2.3.4...1000}=\frac{1.\left(2.3.4....999\right)}{\left(2.3.4....999\right).1000}\)\(=\frac{1}{1000}\)

b) Đặt: A=\(1+2+2^2+2^3+...+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+....+2^{2008}+2^{2009}\)

\(\Leftrightarrow2A-A=2^{2009}-1\)

\(\Leftrightarrow A=2^{2009}-1\)

\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)\(=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=\frac{1}{-1}=-1\)

vậy: S=(-1)