What are your thoughts?Prove nc1+nc2+nc3+...+ncn=2^(n)-1
I have absolutely zero idea on how todo this questionc so any hints would be appreciated. Thanks.
Well I think I’d need to sub a value for x in (1+x)^n, looking at the question I can’t see how deriving or integrating would help. Like I said, I’ve got no clue on how to start going about this apart from what I mentioned.What are your thoughts?
Please share your work with us ...even if you know it is wrong.
If you are stuck at the beginning tell us and we'll start with the definitions.
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Hint: Use
nCr = \(\displaystyle \ \frac{n!}{r!(n-r)!}\)
Can you expandWell I think I’d need to sub a value for x in (1+x)^n, looking at the question I can’t see how deriving or integrating would help. Like I said, I’ve got no clue on how to start going about this apart from what I mentioned.
so the first term would be nc0*a^n*b^0 and in this case a=1 and b=x hence the first term would be 1. Letting x=1 we get 2^n and then bringing the first term to the other side we get the required proof. Thanks for the help.Can you expand
(a+b)n
- using binomial theorem?
I‘ve got it thanks.You might observe that the right-hand side contains 2^n; how about expanding (1+1)^n by the binomial theorem?
Put a = b = 1so the first term would be nc0*a^n*b^0 and in this case a=1 and b=x hence the first term would be 1. Letting x=1 we get 2^n and then bringing the first term to the other side we get the required proof. Thanks for the help.
I‘ve got it thanks.
\(\displaystyle \displaystyle{\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {\dbinom{n}{k}{a^k}{b^{n - j}}}\) letting \(\displaystyle a=b=1\) we get:Prove nc1+nc2+nc3+...+ncn=2^(n)-1