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This is really a simple (pre?)algebra issue, but this type of problem is causing me trouble over and over, so I want to understand and get to the heart of these types of problems.

I'm trying to simplify this equation as part of an inductive proof of a recurrence relation.

$5 \left(2^{k + 1} + 3^k\right) - 6 \left(2^k + 3^{k - 1}\right)$

I can run this through wolfram alpha and see it's equivalent to what I need, which is:


But I'm having trouble figuring out how to get there myself. Can someone break down the steps for me (and/or point me somewhere)? Thanks! [edit: fixed the k-1 to be k+1]

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Expand $6(2^k + 3^{k-1}) = 6 \cdot 2^k + 6 \cdot 3^{k-1}$. Now remember 6 = 2*3. – Ted Apr 7 '12 at 21:49
up vote 2 down vote accepted

Hint $\: $ Put $\rm\:a,b\: =\: 2,3\ \ in\ \ (a+b)\ (\color{red}{a^{k+1}}+b^k) \color{green}{ -ab}\ (\color{brown}{ a^k} + b^{k-1})\: =\ a^{k+2} + b^{k+1}$

which holds true because $\rm \ \ (a+b)\ \color{red}{a^{k+1}}\color{green}{ - ab}\ \color{brown}{a^k}\ =\ ((a+b)\:\color{red}a - \color{green}{ab})\:\color{brown}{a^k} = a\:\!\color{red}a\:\!\color{brown}{a^k} = \color{brown}a^{\color{brown}{k+2}}$

and similarly for powers of $\rm\:b$.

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Here's a bull-headed way of doing it: just expand, group together powers of $2$ and powers of $3$, and compute: $$\begin{align*} 5\Bigl(2^{k+1}+3^k\Bigr) - 6\Bigl(2^k+3^{k-1}\Bigr) &= 5\Bigl( 2^{k+1}\Bigr) + 5\Bigl(3^k\Bigr) - 6\Bigl(2^k\Bigr) - 6\Bigl(3^{k-1}\Bigr)\\ &= (5\times 2 - 6)2^k + (5\times 3 - 6)3^{k-1}\\ &= (10-6)2^k + (15-6)3^{k-1}\\ &= 4\times 2^k + 9\times 3^{k-1}\\ &= 2^2\times 2^k + 3^2\times 3^{k-1} \end{align*}$$ and the final result follows.

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