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$5(2^{n−1} + 5 ·3^{n−1}) − 6(2^{n−2} + 5 · 3^{n−2}) = 2^{n−2}[10 − 6] + 3^{n−2}[75 − 30] = 2^{n−2} · 4 + 3^{n−2} · 9 · 5 = 2^n + 3^n · 5 $

There are enormous leaps in my understand between each section listed. The answer is there, but I don't understand how they arrived at it. Would anyone mind trying to clarify? Been out of College Algebra for quite a few years and I'm taking a upper graduate Discrete Math class that requires this proof for recurrence relations.

Daniel Fischer
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Logan
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    Hint: $5(2^{n-1}) = 5(2^1 (2^{n-2})) = 10(2^{n-2})$. You want to get all the exponents to be the same to simplify. – Amzoti Sep 28 '13 at 17:31
  • I feel like I'm clearly forgetting something integral. How do you get from $2^{n−1}$ to $2^{1}(2^{n−2})$ ? – Logan Sep 28 '13 at 17:54
  • You know that $2^1 \times 2^{-2} = 2^{-1}$, by law of exponents. Clear? So, $2^1(2^{n−2})= 2^{n-2+1} = 2^{n-1}$ – Amzoti Sep 28 '13 at 19:13

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As the lowest power of $2$ is $n-2,$

we set $2^{n-2}=a\implies 2^{n-1}=2\cdot2^{n-2}=2a$

So, if we consider the terms containing the power of $2,$

$5(2^{n−1}) − 6(2^{n−2}) =5(2a)-6(a)=a(5\cdot2-6)=4a=2^2\cdot2^{n-2}=2^{(2+n-2)}=2^n$

Similarly, for the terms containing the power of $3$

lab bhattacharjee
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