Question

Cho C= \(\frac{3}{4}\) + \(\frac{15}{16}\)+......+\(\frac{4^{2010}-1}{4^{2010}}\)

So sánh C với 2009

Answer 1

\(C=\frac{4^1-1}{4^1}+\frac{4^2-1}{4^2}+...+\frac{4^{2009}-1}{4^{2009}}+\frac{4^{2010}-1}{4^{2010}}\)

\(C=\frac{4^1}{4^1}-\frac{1}{4^1}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{4^{2009}}{4^{2009}}-\frac{1}{4^{2009}}+\frac{4^{2010}}{4^{2010}}-\frac{1}{4^{2010}}\)

\(C=\left(1+1+...+1\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)(tổng có 2010 số 1)

\(C=2010-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)

Xét tổng \(A=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\)

=> \(4A=1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\)

=> \(4A-A=\left(1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\right)-\)\(\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)

=> \(3A=1-\frac{1}{4^{2010}}2010-1>2009\)

 

Answer 2

\(C=\frac{4^1-1}{4^1}+\frac{4^2-1}{4^2}+...+\frac{4^{2009}-1}{4^{2009}}+\frac{4^{2010}-1}{4^{2010}}\)

\(C=\frac{4^1}{4^1}-\frac{1}{4^1}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{4^{2009}}{4^{2009}}-\frac{1}{4^{2009}}+\frac{4^{2010}}{4^{2010}}-\frac{1}{4^{2010}}\)

\(C=\left(1+1+...+1\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)(có 2010 số 1)

\(C=2010-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)

Xét : \(A=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\)

\(4A=1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\)

\(4A-A=\left(1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)

\(3A=1-\frac{1}{4^{2010}}< 1\)

\(A< \frac{1}{3}\)

\(C=2010-A>2010-\frac{1}{3}>2010-1>2009\)