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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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    Hint: $(k-1)+(k+1)=2k=2((k+1)-1)$. Multiply all this by $2^{k+1}$ and add $2$.2012-04-02
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    @bgins: thats it!!2012-04-02

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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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    Why this step: $(k-1)2^{k+1} + (k+1)2^{k+1} = \Bigl( (k-1)+k+1)\Bigr)2^{k+1}$2012-04-02
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    What do you mean, "Why this step"? I'm factoring out $2^{k+1}$, which is a common factor to both summands. It's just the distributivity property, $ac+bc = (a+b)c$.2012-04-02
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    thank you. Also why 2k? could you please type it as you typed this: $(ac+bc=(a+b)c)$2012-04-02
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    Are you kidding me?? What is $k-1+k+1$?2012-04-02
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    I am not kidding you, I am really lost when it comes to algebra!!2012-04-02
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    If you cannot see how to go from $k-1+k+1$ to $2k$, then you shouldn't be trying to do induction proofs, you should be going back to grade 7 and learning basic algebra.2012-04-02
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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 $2+(k)2^{k+2}$.