Suppose that:

$\large A = (1 +2017)^{2017} \\ \\ \large B = (1 + 2019)^{2016}$

Then which of the following is true?

A > B
none of them
A < B
A = B

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$\begin{aligned} & A = (1+2017)^{2017} = 2018^{2017} \\ & B=(1+2019)^{2016} = 2020^{2016}\end{aligned}$

Let $p = \frac{A}{2018} = \frac{2018^{2017}}{2018} = 2018^{2016}$ . Then

$\begin{aligned}\frac{p}{B} = \frac{A}{2018B} & = \frac{2018^{2016}}{2020^{2016}} \\ & = 2018\left( \frac{2018}{2020}\right)^{2016} \\& = 2018\left(1-\frac{2}{2020}\right)^{2016 }\\ & = 2018q > 1 \end{aligned}$ .

where $q$ exists in $0<q <1$ Thus $A>B$ .