Đáp án:
$\text{Chúc bạn học tốt}$
Giải thích các bước giải:
$A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+\dfrac{1}{1+3+5+7}+..+\dfrac{1}{1+3+5+..+2017}$
$A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{1018081}$
hay $A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{1009^2}$
Mà $A<\dfrac{1}{4}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+..+\dfrac{1}{1008×1009}$
$ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{1008}-\dfrac{1}{1009}$
$⇒A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}$
$⇒A<\dfrac{3}{4}-\dfrac{1}{1009}$
Vì $\dfrac{3}{4}-\dfrac{1}{1009}<\dfrac{3}{4}$
$⇒A<\dfrac{3}{4}$
Vậy đpcm
