Giải thích các bước giải:
Ta có:
$\dfrac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}$
$=\dfrac{k+2}{k!\left(1+\left(k+1\right)+\left(k+1\right)\left(k+2\right)\right)}$
$=\dfrac{k+2}{k!\left(\left(k+2\right)+\left(k+1\right)\left(k+2\right)\right)}$
$=\dfrac{k+2}{k!\left(k+2\right)\left(1+\left(k+1\right)\right)}$
$=\dfrac{k+2}{k!\left(k+2\right)\left(k+2\right)}$
$=\dfrac{1}{k!\left(k+2\right)}$
$=\dfrac{k+1}{k!\left(k+2\right)\left(k+1\right)}$
$=\dfrac{k+1}{\left(k+2\right)!}$
$=\dfrac{\left(k+2\right)-1}{\left(k+2\right)!}$
$=\dfrac{1}{\left(k+1\right)!}-\dfrac{1}{\left(k+2\right)!}$
Áp dụng công thức trên
$\to P=\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+....+\dfrac1{2016!}-\dfrac{1}{2017!}$
$\to P=\dfrac1{2!}-\dfrac{1}{2017!}$
$\to P<\dfrac1{2!}$
$\to P<\dfrac12$
