Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
 I tried solving but got stuck. 
Show that it is true for n=1
 $$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
 is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
 I'm stuck and don't know how to proceed
 A: Hint:
$24 $ divides $2\cdot7^n+3\cdot5^n-5$ for $n=0$ and $1,$
and $2\cdot7^{n+2}+3\cdot5^{n+2}-5=49(2\cdot7^n+3\cdot5^n-5)-24(3\cdot5^n-10).$
A: Notice that $5^2$ and $7^2$ are both one more than a multiple of $24.$  Prove the cases $k=1$ and $k=2$ for your basis step. 
Then in the induction you have, for one term,
$$2(7^{k+2}) = 2(7^k)(48+1).$$
Do the same for the other term.  Induct by twos instead of by ones.
A: In essence, the induction step relies on the identity $$(ax^n + by^n)(x+y) = (ax^{n+1} + by^{n+1}) + xy(ax^{n-1} + by^{n-1}).$$  You can verify this by simple multiplication.  So if we define $$f_n(a,b,x,y) = ax^n + by^n,$$ then we have the recursion relation $$f_{n+1}(a,b,x,y) = (x+y)f_n(a,b,x,y) - xy f_{n-1}(a,b,x,y).$$  In your case, it is natural to choose $a = 2$, $b = 3$, $x = 7$, $y = 5$ to obtain $$f_{n+1} = 12 f_n - 35 f_{n-1}$$ (where I have omitted the arguments of the function).  So if we let $g_n = f_n - 5$, then $$g_{n+1} + 5 = 12(g_n + 5) - 35(g_{n-1} + 5)$$ or $$g_{n+1} = 12g_n - 35 g_{n-1} - 120.$$  Since $120 = 5(24)$, it follows that $24 | g_{n+1}$ whenever $24 | g_n$ and $24 | g_{n-1}$; since $g_0 = 0$ and $g_1 = 24$, the proof is complete.
A: First note that $5^k-5$ is divisible by $4$ for all positive integers $k$.
Let $P(n)$ be the proposition that $2\times7^n+3\times5^n-5$ is a multiple of $24$.
$P(1)$ is the proposition that $2\times7+3\times5-5$ is a multiple of $24$ which is true
Suppose $P(k)$ is true and so $2\times7^k+3\times5^k-5$ is  a multiple of $24$.
Then $2\times7^{k+1}+3\times5^{k+1}-5=14\times 7^k+15\times5^k-5=7\times (2\times7^k+3\times5^k-5)-6\times(5^k-5)$
is divisible by $24$. Therefore $P(k+1)$ is true and we are finished.
A: If induction is not mandatory,
I believe this is how the problem came into being
$$2(1+6)^n+3(1+4)^n\equiv2(1+6n)+3(1+4n)\pmod{(72,48)}$$
