Question

[SOLVED] Prove that \[\sum_{j=0}^n (-1)^j \binom{n}{j} = 0\]For all \(n > 0\).

Answer 1

I feel like I'm missing some basic trick.

Answer 2

induction?

Answer 3

Supposedly, but I'm not sure how to set it up properly.

Answer 4

I dont see any values to avoid perse

Answer 5

lets create a base case

Answer 6

is it true for n=1?

Answer 7

\[\binom{1}{0} - \binom{1}{1}=1-1=0\]Check.

Answer 8

good, how about 2 and 3 for good measure?

Answer 9

and isnt this just the pascal rows? with alternating signs ....

Answer 10

\[\binom{2}{0} - \binom{2}{1} + \binom{2}{1}=1-2+1=0\]\[\binom{3}{0} - \binom{3}{1} + \binom{3}{2} - \binom{3}{3}=1-3+3-1=0\]Check.

Answer 11

So we can assume it's true for some \(n=k\). My issue is then extending that to \(k+1\).

Answer 12

1-4+6-4+1 = 8-8 = 0

Answer 13

well, lets write up the k proposal wo we have something to wirk with

Answer 14

I see a combinatorics argument we could make depending on the fact that n is odd or even.

Answer 15

i think pascals is proved inductively too isnt it

Answer 16

As for the \(k\) proposal...\[\sum_{j-0}^k (-1)^j \binom{k}{j} = 0\]Show that \[\sum_{j-0}^{k+1} (-1)^j \binom{k+1}{j} = 0\]

Answer 17

yeah, that second one is familiar to the proof for the pascals ....

Answer 18

id have to google it to iron out the details

Answer 19

I see the correct relation to Pascal's rule now. Let's see where that takes us.

Answer 20

\[\sum_{j-0}^{k+1} (-1)^j \binom{k+1}{j} = \sum_{j-0}^{k} (-1)^j \left[\binom{k}{j} + \binom{k}{j-1}\right]\]So we can distribute, and then by our hypothesis, we know that it's true since we're using \(k\) instead of \(k+1\).

Answer 21

How about Expanding \( (1-x)^n\) and then substituting \(x=1\)?

Answer 22

I suppose that might just be a better/easier way to do it.

Answer 23

You suppose? It is actually :)

Answer 24

Maybe I just like the more complicated ways better. :)

Answer 25

Good luck :)