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Write the following sequence as a recurrence relation (with sufficient initial values specified):

$$b_n=1-\frac{1}{2^n} \forall n\in\mathbb{N}^*$$

I think I am suppose to use induction (e.g. n-1) to prove the recurrence, but I am confused, how?

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There are infinite ways to create a recurrence relation. – Aryabhata Feb 16 '11 at 22:42
This is just my homework I copied straight out of the book...I believe it wants the form asubn = asubn-1 + ???. (stating asub1 = ???). – Mr_CryptoPrime Feb 16 '11 at 22:46
So what is $b_n - b_{n-1}$? What is $b_1$ or $b_0$? – Henry Feb 17 '11 at 0:04
up vote 6 down vote accepted

HINT $\ $ Eliminate $\rm\ 1/2^n\ $ from the equations for $\rm\ b_n,\ b_{n+1}\ $

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If you wrote $$\frac{1}{2^{n}} = 1 - b_{n}$$ and similarly for the $n-1$ case then you could probably see a relationship between $b_{n-1}$ and $b_n$

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