# Prove this

Prove that $$\sum_{r=1}^n$$ $$(-1)^{r-1} \times nCr$$ = 1

Note by Selena Miller
4 years, 5 months ago

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## Comments

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How many ways can you prove the identity?

Can you use an algebraic method? Can you use a combinatorial method? Can you find a pattern? Can you relate it to Pascal's Triangle?

Staff - 4 years, 5 months ago

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Sum from 0, use the Binomial expansion for (1-1)^n

- 4 years, 5 months ago

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Solution 1

Expand $$(1-1)^n$$ according to the binomial theorem. We obtain

$$\displaystyle\sum_{r=0}^n (-1)^r \binom{n}{r} = 0$$

$$\displaystyle\sum_{r=0}^n (-1)^{r-1} \binom{n}{r} = 0$$

$$\displaystyle\sum_{r=1}^n (-1)^{r-1} \binom{n}{r} + (-1)^{0-1} \binom{n}{0} = 0$$

$$\displaystyle\sum_{r=1}^n (-1)^{r-1} \binom{n}{r} + (-1) \cdot 1 = 0$$

$$\displaystyle\sum_{r=1}^n (-1)^{r-1} \binom{n}{r} = 1$$

Solution 2

Suppose that there are $$n$$ distinguishable objects numbered $$1,2,\ldots,n$$. We want to pick a non-empty subset of them. Call a subset odd or even according to their cardinality.

Positive terms count the number of ways to choose an odd number of objects (the terms are $$\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots$$). Negative terms count the number of ways to choose an even number of objects (the terms are $$\binom{n}{2}, \binom{n}{4}, \binom{n}{6}, \ldots$$).

Now distinguish object $$n$$. For any subset $$S$$ picked among the remaining $$n-1$$ objects, we obtain two subsets for the $$n$$ objects: $$S$$ and $$S \cup \{n\}$$. One of these is odd and the other is even, so there are the same number of odd subsets as there are even subsets...except that when $$S$$ is empty, $$S$$ is not a valid subset but $$S \cup \{n\}$$ is, so this exception increments the number of odd subsets without incrementing the even one. So there is 1 more odd subset, which is exactly what the identity is.

I'll leave the solution with Pascal's Triangle to others.

- 4 years, 5 months ago

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