Đáp án: $S>3$
Giải thích các bước giải:
Ta có:
$S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2010}{2^{2009}}+\dfrac{2011}{2^{2010}}$
$\to 2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2010}{2^{2008}}+\dfrac{2011}{2^{2009}}$
$\to 2S-S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2008}}+\dfrac{1}{2^{2009}}+\dfrac{2011}{2^{2010}}$
$\to S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2008}}+\dfrac{1}{2^{2009}}+\dfrac{2011}{2^{2010}}$
Đặt $B=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2008}}+\dfrac{1}{2^{2009}}$
$\to 2B=1+\dfrac{1}{2^1}+...+\dfrac{1}{2^{2007}}+\dfrac{1}{2^{2008}}$
$\to 2B-B=1-\dfrac{1}{2^{2009}}$
$\to B=1-\dfrac{1}{2^{2009}}$
$\to S=2+(1-\dfrac{1}{2^{2009}})+\dfrac{2011}{2^{2010}}$
$\to S=2+1-\dfrac{1}{2^{2009}}+\dfrac{2011}{2^{2010}}$
$\to S=3-\dfrac{2}{2^{2010}}+\dfrac{2011}{2^{2010}}$
$\to S=3+\dfrac{2011-2}{2^{2010}}$
$\to S=3+\dfrac{2009}{2^{2010}}$
$\to S>3$
