# Binomial Proof

• October 12th 2010, 06:00 AM
a8281333
Binomial Proof
I havn't done this in a long time! And apparently I should know this easy, it sort of looks like a proof by induction, which I havn't done before and I am frantically trying to learn!

Show that for each integer n the alternating sum of binomial coefficients:

1 - (n) + ... + (-1)^k(n) + ... + (-1)^(n-1)( n ) + (-1)^n
.....(1)....................(k)................... .......(n-1)
is zero. What is the value of the sum

(so what ive done here is started with an "inductive basis of n=1 which kind of suggests it goes to zero but without the appropriate conciseness)

1 + (n) + ... + (n) + ... + ( n ) +1
......(1)...........(k)...........(n-1)

I understand the layout is a bit rubbish but I hope you can fathom it!
Any help would be greatly appreciated!
• October 12th 2010, 06:08 AM
tonio
Quote:

Originally Posted by a8281333
I havn't done this in a long time! And apparently I should know this easy, it sort of looks like a proof by induction, which I havn't done before and I am frantically trying to learn!

Show that for each integer n the alternating sum of binomial coefficients:

1 - (n) + ... + (-1)^k(n) + ... + (-1)^(n-1)( n ) + (-1)^n
.....(1)....................(k)................... .......(n-1)
is zero. What is the value of the sum

(so what ive done here is started with an "inductive basis of n=1 which kind of suggests it goes to zero but without the appropriate conciseness)

1 + (n) + ... + (n) + ... + ( n ) +1
......(1)...........(k)...........(n-1)

I understand the layout is a bit rubbish but I hope you can fathom it!
Any help would be greatly appreciated!

For any two real (or even complex) numbers $a,b:\,\,(a+b)^n=\sum\limits^n_{k=0}\binom{n}{k}a^{ n-k}b^k\,,\,\,n\in\mathbb{N}$ .

You have $\sum\limits^n_{k=0}\binom{n}{k}(-1)^k=\sum\limits^n_{k=0}\binom{n}{k}1^{n-k}(-1)^k=...$ and the result is immediate.

Tonio
• October 12th 2010, 06:22 AM
a8281333
so basically, as k is even then odd, they alternate, basically giving a sequence that cancels out? I hope this is what you mean, I'm really keen to learn the process in detail to build a firm fundamental understanding
• October 12th 2010, 06:28 AM
tonio
Quote:

Originally Posted by a8281333
so basically, as k is even then odd, they alternate, basically giving a sequence that cancels out? I hope this is what you mean, I'm really keen to learn the process in detail to build a firm fundamental understanding

No, I did not mean that, though that is what happens. Do read again CAREFULLY and pay attention to the $(a+b)^n=...$ thing

Tonio
• October 12th 2010, 06:42 AM
a8281333
Sorry, I dont understand, this must be really frustrating for you! I best get my head in a book, I havn't done this sort of mathematics in two years so I'm very rusty, but I'm sure once I discover the answer/steps it will all come rushing back! :)
• October 12th 2010, 06:52 AM
a8281333
after a bit of research, am i right to assume this is a telescoping series?
• October 12th 2010, 07:05 AM
a8281333
Update, apparently I dont need induction? I just expand (1+1)^n and (1-1)^n, that doesnt seem right to me though? anyone care to shread any light on this?
• October 12th 2010, 07:46 AM
TheCoffeeMachine
Quote:

Originally Posted by a8281333
Update, apparently I dont need induction? I just expand (1+1)^n and (1-1)^n, that doesnt seem right to me though? anyone care to shread any light on this?

More like just the second one. I think you didn't pay attention to what Tonio was saying.

The binomial theorem states $\displaystyle \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k = (a+b)^n$

What the sum you have is $\displaystyle \sum_{k=0}^{n}\binom{n}{k}(1)^{n-k}(-1)^k$

Don't you see what $a$ and $b$ are in this case?
• October 12th 2010, 07:49 AM
a8281333
Ah yes! I see, it is just (1-1)^n, then the second part is just 2^n!, how could i be so stupid! I love learning :) so my solution for part 2 is just (1+1)^n which is just 2^n right?! Woop
• October 12th 2010, 07:50 AM
TheCoffeeMachine
Quote:

Originally Posted by a8281333
so my solution for part 2 is just (1+1)^n which is just 2^n right?! Woop

Yes. (Yes)