Binomial Proof

[Afridi]

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.

 

[Deleted member 4993]

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.

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.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

Hint: Use

_nC_r = \(\displaystyle \ \frac{n!}{r!(n-r)!}\)

 

[Afridi]

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.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

Hint: Use

_nC_r = \(\displaystyle \ \frac{n!}{r!(n-r)!}\)

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.

 

[Deleted member 4993]

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.

Can you expand

(a+b)ⁿ

- using binomial theorem?

 

[Dr.Peterson]

You might observe that the right-hand side contains 2^n; how about expanding (1+1)^n by the binomial theorem?

 

[Afridi]

Can you expand

(a+b)ⁿ

- using binomial theorem?

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.

You might observe that the right-hand side contains 2^n; how about expanding (1+1)^n by the binomial theorem?

I‘ve got it thanks.

 

[Deleted member 4993]

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.

I‘ve got it thanks.

Put a = b = 1

 

[pka]

Prove nc1+nc2+nc3+...+ncn=2^(n)-1

\(\displaystyle \displaystyle{\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {\dbinom{n}{k}{a^k}{b^{n - j}}}\) letting \(\displaystyle a=b=1\) we get:
\(\displaystyle \displaystyle{2^n} = \sum\limits_{k = 0}^n {\dbinom{n}{k}}\)
\(\displaystyle \displaystyle{2^n} =1+ \sum\limits_{k = 1}^n {\dbinom{n}{k}}\)
Can you finish?