Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$? I have question 

Q
  Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?

by using induction  
Thanks
 A: Let $f(n)=2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$f(m+1)-f(m)=4\cdot6^{m+1}+5(m+1)-4-(4\cdot6^m+5m-4)$
$=5\{5+4(6^m-1)\}$
Now, we know $(6^m-1)$ is divisible by $6-1=5$
$\implies f(m+1)\equiv f(m)\pmod {25}$
Now, $f(1)=2^3\cdot3+5\cdot1-4=25\implies f(1)$ divisible by $25$

Alternatively,
$2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$=4(1+5)^n-5n-4$
$=4\left(1+\binom n1 5+\binom n2 5^2+\cdots +\binom n{n-1}5^{n-1}+5^n\right)+5n-4$ (Using Binomial Expansion)
$= 4\{1+5n+5^2\left(\binom n2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)\}+5n-4$ 
As the Binomial  coefficients are all integers, all the terms after $\binom n1 5$ is divisible by $5^2=25,$
$5^2\left(\binom n25^2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)$ can bewritten as $25k$ where $k$ is an integer
So, $2^{n+2}3^n+5n-4=4\{1+5n+25k\}+5n-4=25(n+4k)$
A: Proof by Induction Second Step
observe 2^(n+3)*3(n+1)+5(n+1)-4=
        2*2^(n+2)*3*3(n)+5n+5-4= with some algebra
        6(2^(n+2)*3(n)+5n-4)-25n+25
Note that by hypothesis 2^(n+2)*3(n)+5n-4 = 25k and therefore the expression can be written as
        25(6k-n+1) which is clearly divisible by 25
