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I am working with mathematical induction, but it gets harder when it comes to convert (or change) the form of the equation with algebra.

I have: $2+(k-1)2^{k+1} + (k+1)2^{k+1}$

And want it to reach this form: $2+((k+1)-1)2^{(k+1)+1}$

What are algebra rules/steps or simplification rules/steps I can use to reach the required form?

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    @bgins: thats it!!2012-04-02

2 Answers 2

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Basic algebra: $\begin{align*} (k-1)2^{k+1} + (k+1)2^{k+1} &= \Bigl( (k-1)+(k+1)\Bigr)2^{k+1} \quad\text{(distributivity of }\times\text{ over }+\text{)}\\ &= 2k\cdot 2^{k+1} \quad\text{(performing the operation)}\\ &= k(2^12^{k+1})\quad\text{(commutativity and associativity of }\times\text{)}\\ &= k2^{1+k+1}\quad\text{(}2^a2^b=2^{a+b}\text{)}\\ &= (k+0)2^{(k+1)+1}\quad\text{(}x+0=x\text{)}\\ &= \Bigl(k+(1-1)\Bigr) 2^{(k+1)+1}\quad\text{(}a-a=0\text{)}\\ &= \Bigl( (k+1)-1\Bigr)2^{(k+1)+1}\quad\text{(associativity of }+\text{)}. \end{align*}$

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    I am doing what are you saying, I am working by myself with KhanAcademy and their Algebra videos :)2012-04-02
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If you want to verify that your equation holds, a reasonable strategy is to try to express each side as "simply" as possible. So let us work separately with the left-hand side and the right-hand side, while glancing at each looking for commonalities.

Left-hand side: The parts $(k-1)2^{k+1}$ and $(k+1)2^{k+1}$ have a common factor $2^{k+1}$. So their sum is $(2k)2^{k+1}$, and the left-hand side is equal to $2+(2k)2^{k+1}$, which can be rewritten as $2+(k)2^{k+2}$.

Right-hand side: Just doing the arithmetic gives us $2+(k)2^{k+2}$.