- #1

Robb

- 225

- 8

## Homework Statement

prove the summation by counting in a set two ways.

**Mod edit: The summation was later confirmed by the OP to be**##\sum_{k = 1}^n 2^{k - 1}##

$$\sum_{n=0}^k 2^{k-1} = 2^{k}-1$$

## Homework Equations

## The Attempt at a Solution

[/B]

##2^{k-1} = (1+1)^{k-1} = \binom n 0 + \binom n 1 + \binom n 2 + ...+ \binom n n = 2^n##

I assume that if I can show that the LHS = RHS = 2^n, then I have proven they are equivalent? I'm not sure how to deal with the RHS because of the (-1). Using the binomial theorem should I say a = 2^k and b = -1 or do I say (1+1)^{k} - 1, where a = 1 and b = 1. Still not sure how to deal with (-1). Also, I have used the binomial theorem on 2^{k-1} so do I need to use a different method on the RHS. Please advise.

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