Use the binomial theorem to Prove: The sum as k=0 --> n OF (nCk)*2^k = 3^n Thanks...
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So you want to prove that $\displaystyle \displaystyle\sum_{k=0}^{n}{n\choose k} 2^{k}=3^{n},$ right? Have you had any ideas so far?
Yes thats correct. And no, I don't see what to do...
Originally Posted by jzellt And no, I don't see what to do... Hint: 1 to any power is 1, i.e. $\displaystyle 1 = (1)^{n-k}$.
I just really don't see this one. Is it possible for someone to show me the proof? Thanks
$\displaystyle (a+b)^n=\displaytype{\sum_{k=0}^n \binom{n}{k}b^ka^{n-k}}$ consider a=1 and b=2
Last edited by BAdhi; Mar 2nd 2011 at 08:56 PM. Reason: latex error
I think I can go from here... Thanks!
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