Đáp án: $T<3$
Giải thích các bước giải:
Ta có:
$T=\dfrac22+\dfrac3{2^2}+\dfrac4{2^3}+...+\dfrac{2019}{2^{2018}}+\dfrac{2020}{2^{2019}}$
$\to 2T=2+\dfrac3{2}+\dfrac4{2^2}+...+\dfrac{2019}{2^{2017}}+\dfrac{2020}{2^{2018}}$
$\to 2T-T=2+\dfrac12+\dfrac1{2^2}+...+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T=2+\dfrac12+\dfrac1{2^2}+...+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
Mà $A=\dfrac12+\dfrac1{2^2}+...+\dfrac{1}{2^{2018}}$
$\to 2A=1+\dfrac12+...+\dfrac1{2^{2017}}$
$\to 2A-A=1-\dfrac{1}{2^{2018}}$
$\to A=1-\dfrac{1}{2^{2018}}$
$\to T=2+1-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T=3-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T<3$
