Question

[naruto]

So sánh tổng S= 1/2+2/2^2+3/2^3+...+n/2^n+...+2007/2^2007 với 2 .( n€n*)

Answer 1

[I Love Family]

\(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}+...+\frac{2007}{2^{2007}}\)

Ta có: \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)

\(\Rightarrow\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}\)

\(=\frac{1}{2}+\left(\frac{3}{2}-\frac{4}{2^3}\right)+\left(\frac{4}{2^3}-\frac{5}{2^3}\right)+...+\left(\frac{2008}{2^{2006}}-\frac{2009}{2^{2007}}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\frac{2009}{2^{2007}}\)

\(=2-\frac{2009}{2^{2007}}< 2\)

~ Học tốt ~ K cho mk nhé! Thank you.