Binomial Theorem - Proof

**mybrohshi5** · February 3rd 2011, 01:30 PM

Show that, for n > 0,

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

I know i am supposed to use the binomial theorem which is just,

$(a+b)^n = \sum_{i=0}^{n}\binom{n}{i} a^i b^{n-i}$

I am really just not sure where to start though

Thank you for any help/tips

(if this needs to be moved to Discrete Mathematics, Set Theory and Logic, would a mod please move it. thank you)

**TheCoffeeMachine** · February 3rd 2011, 01:34 PM

Hint: $\displaystyle \sum_{i=0}^{n}\binom{n}{i}(-1)^i = \sum_{i=0}^{n}\binom{n}{i}(-1)^i(1)^{n-i}.$

**Plato** · February 3rd 2011, 01:52 PM

Here is a different hint.
$0=(1-1)^n=~?$ That is of course a sum.

**Archie Meade** · February 3rd 2011, 02:50 PM

Or, you could let

$a=b=-\frac{1}{2}$

or something similar and plug them into the expansion, noticing a pattern.

**mybrohshi5** · February 3rd 2011, 05:36 PM

Ok great i think i got it.

for a = 1 and b = -1 and using any number where n > 0 you can see that when plugged into the binomial theorem it is a repeating pattern of (in this case) 2 + -2 + 2 + -2 .... which therefore will = 0 proving $\sum_{i=0}^{n}(-1)^i\binom{n}{i} = 0$ to be true.

Does this seem correct or am i missing something? I am very new to proofs (this is actually my first one haha

)

**Archie Meade** · February 4th 2011, 03:49 AM

Originally Posted by mybrohshi5

Ok great i think i got it.

for a = 1 and b = -1 and using any number where n > 0 you can see that when plugged into the binomial theorem it is a repeating pattern of (in this case) 2 + -2 + 2 + -2 .... which therefore will = 0 proving $\sum_{i=0}^{n}(-1)^i\binom{n}{i} = 0$ to be true.

Does this seem correct or am i missing something? I am very new to proofs (this is actually my first one haha

)

Why would you get 2+(-2)+2+(-2)+.... ?

What would the $\binom{n}{i}$ values be ?

and "yes", you are missing something, not only that n could be even giving an odd number of terms,
so the sum would not be zero.

TheCoffeeMachine and Plato are drawing your attention to the fact that

$\displaystyle\ (1-1)^n=\binom{n}{0}(-1)^01^n+\binom{n}{1}(-1)^11^{n-1}+\binom{n}{2}(-1)^21^{n-2}+.....+\binom{n}{n-1}(-1)^{n-1}1^1+\binom{n}{n}(-1)^n1^0$

The expansion is $\displaystyle\sum_{i=0}^n\binom{n}{i}(-1)^i1^{n-i}$

and $1-1=0$

**mybrohshi5** · February 6th 2011, 04:25 PM

I have been looking over these posts again for the past day or so and I still can't seem to wrap my mind around this.

I get using A = -1 and B = 1 and using the binomial theorem will give you the expansion

$\displaystyle\ (-1 + 1)^n=\binom{n}{0}(-1)^01^n+\binom{n}{1}(-1)^11^{n-1}+\binom{n}{2}(-1)^21^{n-2}+.....+\binom{n}{n-1}(-1)^{n-1}1^1+\binom{n}{n}(-1)^n1^0$

I guess i just dont get how this proves that for n > 0

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

I feel like i am making this much more difficult than it should be

**Archie Meade** · February 6th 2011, 04:44 PM

It's because

$1^{n-i}=1$

in each term of the expansion.

TheCoffeeMachine related the binomial expansion to your closed-form sum.

They are the same.
Hence there is no need to calculate each term of the expansion, since the sum is

$(1-1)^n=0^n=0$

as Plato also showed.

**mybrohshi5** · February 6th 2011, 04:55 PM

Thank you for the great explanation

I am going to give my brain a rest then rework the proof and make sure i understand everything and if I dont I will post back again

Thanks again for all the replies everyone. I really appreciate the help

**Krahl** · February 6th 2011, 06:43 PM

You said you are supposed to use the binomial theorem. So for $n > 0$ we have,
$(a+b)^n = \sum_{i=0}^{n}\binom{n}{i} a^i b^{n-i}$
let a = -1 and b = 1. This equality then becomes,
$((-1)+(1))^n = \sum_{i=0}^{n}\binom{n}{i} (-1)^i (1)^{n-i}$
re-written,
$(0)^n = \sum_{i=0}^{n}\binom{n}{i} (-1)^i$

So the binomial theorem actually proves it for you. You just need to plug in those values. This is just a special case of the binomial theorem.

You could also use the binomial theorem to show that
$((1)+(1))^n = \sum_{i=0}^{n}\binom{n}{i} (1)^i (1)^{n-i}$
So that
$(2)^n = \sum_{i=0}^{n}\binom{n}{i}$

in each case you are using the special cases of a theorem that has already been proven for you. I hope that helps you out a bit.

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