$$\sum_{k=0}^{n}\binom nk=2^n$$ then how is $2(\binom n0+\binom n2+\binom n4+...)=2^n$ ?? I don't think it could be because half of the members of the sum are chosen, that seems a bit intuitively risky to conclude, what am i missing here ?
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2The even terms have the same sum as the odd terms. – vadim123 Apr 17 '15 at 15:09
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How is this possible ?? – Bozo Vulicevic Apr 17 '15 at 15:10
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I cannot believe this is not a duplicate. Time to search. – Jack D'Aurizio Apr 17 '15 at 15:20
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We have $$(1+x)^n = \sum_{k=0}^n \dbinom{n}k x^k$$ Plugging in $x=1$, we obtain $$2^n = \sum_{k=0}^n \dbinom{n}k$$ Plugging in $x=-1$, we obtain $$0 = \sum_{k=0}^n \dbinom{n}k (-1)^k$$ Adding both we obtain what you want.
Adhvaitha
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$\sum_{k=0}^{n}\binom nk=2^n$ results from substituting $x = 1$ in the binomial expansion $(1+x)^{n}=\sum_{k=0}^{n}\binom nk x^k$. Just put $x = -1$ in the same expansion to get $\sum_{k=0}^{n}(-1)^k\binom nk = 0$. Adding these two relations cancels out the odd terms and you get the desired result.
Seven
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