# Show that $7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$

$7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$

The problem is part of a proof. If you could also talk me through your thought process for solving this problem, I would greatly appreciate it. I've played around with several things like trying to get everything in terms of $a^{k+1}$ but I'm having no luck. Thanks!

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your last bracket is ) instead of } – Dominic Michaelis Feb 19 '13 at 22:35
Looks like you may need to fix your curly brackets. That should take care of the latex, but I think you may be missing some parentheses as well. – Mike Feb 19 '13 at 22:35

$$7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) =\\21\cdot 2^k+14\cdot 5^k-30\cdot 2^{k-1} -20\cdot 5^{k-1}=\\ 21\cdot 2^k +14\cdot 5^k-15\cdot 2^k-4\cdot 5^k=\\ 6\cdot 2^k+10\cdot 5^k=\\3(2)^{k+1} + 2(5)^{k+1}$$
Hint $\$ Note that $\rm\,\color{#C00}{2}^k,\, \color{#0A0}5^k\:$ are solutions of the linear recurrence $\rm\:f(k\!+\!1) - 7 \ f(k) + 10\ f(k\!-\!1) = 0\:$ hence so too are all linear combinations $\rm\:c\, 2^k + d\, 5^k,\:$ for $\rm\:c,d\:$ constants (independent of $\rm\:k$).
If you know a little about recurrences you can make it even easier by noting that the characteristic polynomial is $\rm\ x^2-7x +10 = (x-\color{#C00}2)(x-\color{#0A0}5).$