# Mathematical Induction, Step after base case.. [closed]

I need to prove the following by mathematical Induction. I have the base case where n=1, and that hold true. However, the step after this have been confusing me. Any help would be great.

## closed as off-topic by Shaun, Claude Leibovici, JonMark Perry, A. Goodier, steven gregoryMar 25 '18 at 12:56

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• You do know the "step after this" is called the induction step? – hardmath Mar 25 '18 at 1:02
• I imagine it's not this particular exercise but rather induction in general which is confusing you. You should consult a book or some online resource. – JKEG Mar 25 '18 at 1:04
• This question might as well be titled "explain induction to me." And you can find about 1000 places where that is answered satisfactorily. – JKEG Mar 25 '18 at 1:06

Using $$(1)\quad\frac 12 + \frac 1{2^2} + \dots+ \frac 1{2^k} = \frac{2^k-1}{2^k}$$ you have to show that, $$(2)\quad\frac 12 + \frac 1{2^2} + \dots+ \frac 1{2^k} + \frac{1}{2^{k+1}}= \frac{2^{k+1}-1}{2^{k+1}}$$ First add $\dfrac{1}{2^{k+1}}$ to (1) to obtain $$\frac 12 + \frac 1{2^2} + \dots+ \frac 1{2^k}+ \frac{1}{2^{k+1}} = \frac{2^k-1}{2^k}+ \frac{1}{2^{k+1}}$$ Now, can you turn $\dfrac{2^k-1}{2^k}+ \dfrac{1}{2^{k+1}}$ into the RHS of (2) ?

• In for a penny, in for a pound. I don't think you'd be cheating the OP by filling in that last step, and conceivably it will help future Readers. Use your own judgement; I'll go ahead and upvote your Answer. – hardmath Mar 25 '18 at 3:19

Guide:

You assume that you know that

$$\frac12+\frac1{2^2}+\ldots + \frac1{2^k}=\frac{2^k-1}{2^k}$$

for a fixed $k$. Use that information as a tool to prove the following:

$$\frac12+\frac1{2^2}+\ldots + \frac1{2^k}+\frac{1}{2^{k+1}}=\frac{2^{k+1}-1}{2^{k+1}}$$

Start from the left hand side and try to get the right hand side.

$$\left(\frac12+\frac1{2^2}+\ldots + \frac1{2^k}\right)+\frac{1}{2^{k+1}}=\ldots$$

Substitute what you know for the part that I have specially put into the brace and simplify the expression.