I have to prove that $\displaystyle 2^{n + 2} \cdot 3^n + 5n - 4$ is divisible by $\displaystyle 25$ for $\displaystyle n>0$.

My proof:

$\displaystyle \begin{array}{l}

2^{n + 2} \cdot 3^n + 5n - 4 = 2^n \cdot 2^2 \cdot 3^n + 5n - 4 = 6^n \cdot 4 + 5n - 4 = \\

5n - 4(6^n - 1) \\

\end{array}

$

This expression is divisible by 25 for $\displaystyle n=1$, so we look for $\displaystyle n>1$.

$\displaystyle 6^n $ is always number which last two digits will be $\displaystyle 36,16,96,76,56,36...$.

So we can establish now for which value of n will last two digits be for number $\displaystyle 6^n$.

$\displaystyle k>0$

$\displaystyle \begin{array}{l}

n = 5k - 3,{\rm{ }}6^{5k - 3} = ....36 \\

n = 5k - 2,{\rm{ }}6^{5k - 2} = ....16 \\

n = 5k - 1,{\rm{ }}6^{5k - 1} = ....96 \\

n = 5k,{\rm{ }}6^{5k} = ....76 \\

n = 5k + 1,{\rm{ }}6^{5k + 1} = ....56 \\

\end{array}

$

Since we have expression $\displaystyle 4(6^n - 1)$ then last two digits of that expression will be:

$\displaystyle \begin{array}{l}

n = 5k - 3,{\rm{ }}4(6^{5k - 3} - 1) = ....40 \\

n = 5k - 2,{\rm{ }}4(6^{5k - 2} - 1) = ....60 \\

n = 5k - 1,{\rm{ }}4(6^{5k - 1} - 1) = ....80 \\

n = 5k,{\rm{ }}4(6^{5k} - 1) = ....00 \\

n = 5k + 1,{\rm{ }}4(6^{5k + 1} - 1) = ....20 \\

\end{array}

$

If we now look at whole expression we will get:

$\displaystyle \begin{array}{l}

n = 5k - 3,{\rm{ }}5(5k - 3) - 4(6^{5k - 3} - 1) \\

n = 5k - 2,{\rm{ }}5(5k - 2) - 4(6^{5k - 2} - 1) \\

n = 5k - 1,{\rm{ }}5(5k - 1) - 4(6^{5k - 1} - 1) \\

n = 5k,{\rm{ }}5(5k) - 4(6^{5k} - 1) \\

n = 5k + 1,{\rm{ }}5(5k + 1) - 4(6^{5k + 1} - 1) \\

\end{array}

$

For example lets take first expression too see if its divisible by 25, same proof is for other ones.

$\displaystyle 5(5k - 3) - 4(6^{5k - 3} - 1) = 25k - 15 - .....40$

.....40 means that last two digits of that number is 40. .....40 - 15 have last two digits 25, so we have number $\displaystyle 25k + .....25$ which is divisible by 25. Same proof is for other numbers.

Is this proof valid?