Tính nhanh giá trị các biểu thức: A= 1/2 + 1/2^2 +1/2^3 +...+1/2^2009 +1/2^2010

Ta có:

\(\frac{2009\ -\frac{2009}{2001}-\frac{2009}{2002}-\frac{2009}{2003}-\frac{2009}{2004}}{2010-\frac{2010}{2001}-\frac{2010}{2002}-\frac{2010}{2003}-\frac{2010}{2004}}=\frac{2009\left(1-\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)}{2010\ \left(1-\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)}=\frac{2009}{2010}\)

Tương tự:

\(\frac{2009\ -\frac{2009}{2005}-\frac{2009}{2006}-\frac{2009}{2007}-\frac{2009}{2008}}{2010-\frac{2010}{2005}-\frac{2010}{2006}-\frac{2010}{2007}-\frac{2010}{2008}}=\frac{2009\left(1-\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)}{2010\ \left(1-\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)}=\frac{2009}{2010}\)

=> \(B=\frac{2009}{2010}:\frac{2009}{2010}=1\)

đề phải là 2²⁰⁰⁹  và 2²⁰¹⁰ mới đúng

Ta có

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)

=> \(2A-A=1-\frac{1}{2^{2010}}\)

=> \(A=\frac{2^{2010}-1}{2^{2010}}\)