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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 $2$. –  bgins Apr 2 '12 at 15:42
    
@bgins: thats it!! –  MIH1406 Apr 2 '12 at 15:44

2 Answers 2

up vote 1 down vote accepted

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: $(k-1)2^{k+1} + (k+1)2^{k+1} = \Bigl( (k-1)+k+1)\Bigr)2^{k+1}$ –  MIH1406 Apr 2 '12 at 15:48
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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, $ac+bc = (a+b)c$. –  Arturo Magidin Apr 2 '12 at 15:49
    
thank you. Also why 2k? could you please type it as you typed this: $(ac+bc=(a+b)c)$ –  MIH1406 Apr 2 '12 at 15:54
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Are you kidding me?? What is $k-1+k+1$? –  Arturo Magidin Apr 2 '12 at 15:56
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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. –  Arturo Magidin Apr 2 '12 at 15:58

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 $2+(k)2^{k+2}$.

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