Proof by induction that $\sum$ $n\choose k $ $= 2^n$

Question

Trying to prove by induction that

$\sum^{i=n}_{i = 0}$$n\choose i$$=2^n$

So obviously, I prove for some arbitary number, i.e $n = 2$

I then go on to show what the n = k terms look like for the first 3 terms

Then I realise to show that it is the same for $n = k+1$, I need to show that:

$2\times(n=k)\equiv (n=k+1)$

Yet when I try and work out the algebra I keep coming excruciatingly close but to no success, am I going about this proof the wrong way or? I've redone my calculations multiple times so there are no errors in my algebraic simplification.

Could it be that I need to go about this another way instead of trying to show it is $2\times n=k$

Use $\binom{n}i=\binom{n-1}i+\binom{n-1}{i-1}$. – Angina Seng Oct 15 '17 at 18:40 — Angina Seng, Oct 15 '17 at 18:40

score 2 · Accepted Answer · answered Oct 15 '17 at 18:38

2

Use the binomial theorem: $$2^n=(1+1)^n=\sum_{i=0}^n\binom{n}{i}1^i1^{n-i}=\sum_{i=0}^n\binom{n}{i}$$

answered Oct 15 '17 at 18:38

Michael Rozenberg

203,855

this is really interesting! can you point me to some sources showing me why what you have written is the case. I never thought about using binomial theorem! – user36606 Oct 15 '17 at 18:43
@user36606 See here: https://en.wikipedia.org/wiki/Binomial_theorem – Michael Rozenberg Oct 15 '17 at 18:45
literally just remembered that the choose operation is used to find binomial coefficients, thank you! – user36606 Oct 15 '17 at 18:46
1

I won't downvote, but this surely is not a proof by induction – b00n heT Oct 15 '17 at 18:47
@b00nheT not this is in particular, he's just showing me what I'd use, which is wanted i wanted to know – user36606 Oct 15 '17 at 18:47
@user36606 You are welcome! – Michael Rozenberg Oct 15 '17 at 18:48
@b00n heT Thank you! It seems that we just had to prove it. – Michael Rozenberg Oct 15 '17 at 18:51

Proof by induction that $\sum$ $n\choose k $ $= 2^n$

1 Answers1