Show that if n is a positive integer, then $${2n\choose0} + {2n\choose2}+ \dots + {2n\choose2n} = 2^{2n-1}$$ I assume the proof is done by induction but how to I do the k+1 case?
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Are you talking about the Lagrange-symbol? Just to be sure. Or are they binomials formatted as fractions? – Ragnar Dec 27 '13 at 22:11
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How is dividing by zero possible? See the first term (2n/0). – babbupandey Dec 27 '13 at 22:11
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What do you mean with 'the $k+1$ case'? – Ragnar Dec 27 '13 at 22:20
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It is a combinatorics questions that is the notation for combinations. – Username Unknown Dec 27 '13 at 22:30
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K+1 case is for induction. – Username Unknown Dec 27 '13 at 22:30
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Suppose you want to take an even number of books out of $2n$ books with you on your holiday. The left hand side sums the number of possibilities over all possible amounts $2k$ to take with you. On the right hand side, for the first $2n-1$ books, you decide whether or not you take it with you. Then the last book is chosen only if you selected an odd number of books within the first $2n-1$ books. Now we have counted the same number of possibilities in two ways, resulting in $$\sum_{k=0}^n\binom{2n}{2k}=2^{2n-1}$$
Ragnar
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Ok am I suppose you use this to prove the equality for the n+1 case? – Username Unknown Dec 27 '13 at 22:31
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No, I don't use induction to prove the equality. this is just a direct proof that is valid for all positive integers $n$. – Ragnar Dec 27 '13 at 22:36
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I dont see how you can do this as a direct proof can you please explain the process? – Username Unknown Dec 27 '13 at 22:38
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My explanation is valid for any $n$, and because we count the same number of possibilities in two ways, the expressions that belong to the two countings have to be equal. – Ragnar Dec 27 '13 at 22:39
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Do you understand all steps I make? This IS the proof, but you may not be used to such a proof. – Ragnar Dec 27 '13 at 22:47
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Apply Pascal's identity and then the binomial theorem. Alternatively, look at the expansion of $\frac{(1+x)^{2n} + (1-x)^{2n}}{2}$ and then plug in $x=1$.
Brian Burns
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The left side counts the number of subsets of a set with 2n elements that have an even number of elements, and there is a bijection between the subsets with an even number of elements and the subsets with an odd number of elements. (See Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized)
Since a set with 2n elements has $2^{2n}$ subsets, the left side must equal $2^{2n-1}$.
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Consider that the sum of a row of Pascal's Triangle behaves this way:
$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+\cdots+\dbinom{2n}{2n}=2^{2n}$
Also that:
$\dbinom{2n}{0}-\dbinom{2n}{1}+\dbinom{2n}{2}-\dbinom{2n}{3}+\cdots+(-1)^{2n}\dbinom{2n}{2n}=0$
Therefore:
$\dbinom{2n}{0}+\dbinom{2n}{2}+\cdots+\dbinom{2n}{2n}= \dbinom{2n}{1}+\dbinom{2n}{3}+\cdots+\dbinom{2n}{2n-1}$
And now you can conclude that:
$2[\dbinom{2n}{0}+\dbinom{2n}{2}+\cdots+\dbinom{2n}{2n}]= 2^{2n}$
Finally, as stated:
$\dbinom{2n}{0}+\dbinom{2n}{2}+\cdots+\dbinom{2n}{2n}=2^{2n-1}$