<s>[INST] 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
 [/INST] 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.
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