Which cancellation should I use?

$\begin{aligned} \left (\dfrac {2015! + 2016!}{2015! \cdot 2016!} \right) \cdot 2016! & = \dfrac {2015!+2016!}{2015!} \\ & = 1 + \dfrac {2016!}{2015!}\\ & = 1 + 2016\\ & = 2017 \end{aligned}$

Let $2015! = a , 2016! = 2016a$

So the expression becomes: $\dfrac{a+2016a}{2016a^2} \times 2016a =\dfrac{2017a}{a} = \huge\boxed{2017}$

$\frac{2015!\times 2017\times 2016!}{2015!\times 2016!}$

=2017

$\frac{2015! + 2016!}{2015! 2016!} 2016!$

Simplify by multiplying $2016!$ and $\frac{1}{2016!}$ , cancelling each other. The new equation would look like this: $\frac{2015! + 2016!}{2015!}$ Factoring $2015!$ from the numerator would give: $\frac{2015! \left( 1 + 2016 \right)}{2015!}$ Simplifying, we'll get $1 + 2016 = \boxed{2017}$

Solution:

$\dfrac{2015! + 2016!}{2015! * 2016!} * 2016!$ = $\dfrac{2015! + 2016!}{2015!}$ = $\dfrac{2015!}{2015!} + \dfrac{2016!}{2015!}$ = $1 + 2016$ = $\boxed{2017}$

First look at $2015!+2016!=2015!+2016\times 2015!=2015!(1+2016)=2017\times 2015!$

so $\frac{2015!+2016!}{2015!\times2016!}\times 2016! =\frac{2017\times2015!}{2015!\times2016}\times2016!=2017$

((2015!+2016!)/(2015!.2016!)).2016!=((2015!+2016!)/2015!) =1+(2016!/2015!)=1+(2016.2015!/2015!)=1+2016=2017

(2015!(1+2016))/(2015!.2016!)*2016!=2017

( (2015 ! +2016 ! )/2015 ! * 2016 ! )(2016 ! )

(2015 ! (1+2016)/2015 ! * 2016!)(2016 ! )

(2015 !) (2017)/2015 !

2017

The 2016! in the denominator and the 2016! outside the fraction will cancel off, leaving you (2015!+2016!)/2015!. You can treat the new fraction as (2015!/2015!)+(2016!/2015!). The first fraction tends to 1, while 2016! can be expressed as 2016x2015!, making the second fraction (2016x2015!)/2015!. Another cancellation will give you 2016. Adding 1 and 2016 will give you 2017.

(2015! (1+2016)/2015!.2016!) 2016!=(2015!.2017/2015!.2016!) 2016!= 2017

The given questions Turns into 2015!/2015!+2016!/2015! =1+2016!/2015! =1+2016 =2017

Let 2!=2015! & 3!=2016! By solving it I got the answer 4..It means that whenever you try to solve two consecutive numbers you will get the next number Therefore my answer is 2017

= 2015!/2015! + 2016 × 2015!/2015! = 1 + 2016 = 2017

((2015)!x(2016+1))/(2015!x2016!)x 2016! 2017

[(2015! + 2016!) ÷ (2015! × 2016!)] × 2016!

= (2015! + 2016!) ÷ 2015!

= 1 + (2016! ÷ 2015!)

= 1 + 2016 ☞(n + 1)! ÷ n! = n+1

= 2017

$\begin{aligned}\left(\dfrac {2015! + 2016!}{2015! \cdot 2016!}\right) \cdot 2016! & = \dfrac {2015! + 2016!}{2015!}\\ & = 1 + 2016\\ & = 2017 \end{aligned}$

(2015!+2015!.2016)/2015! 2015!(1+2016)/2015! (1+2016)=2017

(2015!(1+2016))/(2015! 2016!) 2016! =2017/2016!*2016! =2017

Note that if we divide every term of the numerator and denominator by 2016!, we obtain 2017. Proof: Let 2015!=a , 2016!=b be. Then ([(a/b) + (b/b)] (b/b))/((a/b) (b/b)). Now we must know that a/b=1/2016 and b/b=1. If we calcule it, we'll obtain 2017.

Mental math. Factored numerator and canceled 2015! and 2016! which leaves 2017