KingGeorge
  • KingGeorge
[SOLVED] Prove that \[\sum_{j=0}^n (-1)^j \binom{n}{j} = 0\]For all \(n > 0\).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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KingGeorge
  • KingGeorge
I feel like I'm missing some basic trick.
amistre64
  • amistre64
induction?
KingGeorge
  • KingGeorge
Supposedly, but I'm not sure how to set it up properly.

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More answers

amistre64
  • amistre64
I dont see any values to avoid perse
amistre64
  • amistre64
lets create a base case
amistre64
  • amistre64
is it true for n=1?
KingGeorge
  • KingGeorge
\[\binom{1}{0} - \binom{1}{1}=1-1=0\]Check.
amistre64
  • amistre64
good, how about 2 and 3 for good measure?
amistre64
  • amistre64
and isnt this just the pascal rows? with alternating signs ....
KingGeorge
  • KingGeorge
\[\binom{2}{0} - \binom{2}{1} + \binom{2}{1}=1-2+1=0\]\[\binom{3}{0} - \binom{3}{1} + \binom{3}{2} - \binom{3}{3}=1-3+3-1=0\]Check.
KingGeorge
  • KingGeorge
So we can assume it's true for some \(n=k\). My issue is then extending that to \(k+1\).
amistre64
  • amistre64
1-4+6-4+1 = 8-8 = 0
amistre64
  • amistre64
well, lets write up the k proposal wo we have something to wirk with
KingGeorge
  • KingGeorge
I see a combinatorics argument we could make depending on the fact that n is odd or even.
amistre64
  • amistre64
i think pascals is proved inductively too isnt it
KingGeorge
  • KingGeorge
As for the \(k\) proposal...\[\sum_{j-0}^k (-1)^j \binom{k}{j} = 0\]Show that \[\sum_{j-0}^{k+1} (-1)^j \binom{k+1}{j} = 0\]
amistre64
  • amistre64
yeah, that second one is familiar to the proof for the pascals ....
amistre64
  • amistre64
id have to google it to iron out the details
KingGeorge
  • KingGeorge
I see the correct relation to Pascal's rule now. Let's see where that takes us.
KingGeorge
  • KingGeorge
\[\sum_{j-0}^{k+1} (-1)^j \binom{k+1}{j} = \sum_{j-0}^{k} (-1)^j \left[\binom{k}{j} + \binom{k}{j-1}\right]\]So we can distribute, and then by our hypothesis, we know that it's true since we're using \(k\) instead of \(k+1\).
anonymous
  • anonymous
How about Expanding \( (1-x)^n\) and then substituting \(x=1\)?
KingGeorge
  • KingGeorge
I suppose that might just be a better/easier way to do it.
anonymous
  • anonymous
You suppose? It is actually :)
KingGeorge
  • KingGeorge
Maybe I just like the more complicated ways better. :)
anonymous
  • anonymous
Good luck :)

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