Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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
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
Are you kidding me?? What is $k-1+k+1$? – Arturo Magidin Apr 2 '12 at 15:56
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 $2+(k)2^{k+2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.