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

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

by using induction

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Answer 1

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 $25$

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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, $2^{n+2}3^n+5n-4=4\{1+5n+25k\}+5n-4=25(n+4k)$