Binomial (1 Viewer)

Omnipotence · Jun 28, 2011

Consider the expansion of (1+x)^2n when x=1, prove that 2n'C'0 + 2n'C'1 + 2n'C'2 + ... + 2n'C'2n = 2^2n-1 + (2n)!/2(n!)^2

Drongoski · Jun 28, 2011

Omni - sorry I've to rush off.

But is the expression correct?

2^2n = 2^(2n-1) + 2^(2n-1 ) and 2^(2n-1) does not equal (2n)!/[2(n!)^2]. You can test using n = 1 0r 2.

Drongoski · Jun 28, 2011

Asked to prove:

$\sum _{i=0}^n \binom {2n}i = 2^{2n-1} + \frac {(2n)!}{2 (n!)^2}$

Proof

$(1+x)^{2n} = \sum _{i=0} ^{2n }\binom {2n}i x^i \implies (1+1)^{2n} = 2^{2n} = \sum_{i=0} ^{2n} \binom {2n} i$

$= \sum_{i=0}^n \binom {2n}i + \sum _{i=n+1}^ {2n} \binom {2n}i$

Now

$\sum_{i=n+1}^{2n} \binom {2n} i = \binom {2n}{n+1} + \binom {2n}{n+2} + \binom {2n}{n+3} + \dots + \binom{2n}{2n-1} + \binom {2n}{2n}$

$= \binom {2n}{n-1} + \binom {2n}{n-2} + \binom {2n}{n-3} + \dots + \binom {2n}{1} + \binom {2n}{0} + \binom {2n}{n} - \binom {2n}{n}$

$= \sum_{i=0}^n \binom {2n}i - \frac {(2n)!}{n! \times n!}$

$\therefore 2^{2n} = 2 \times \sum_{i=0}^n \binom {2n} i - \frac {(2n)!}{(n!)^2}$

$\therefore \sum_{i=0} ^{n} \binom {2n} i = 2^{2n-1} + \frac {(2n)!}{2 \times (n!)^2}$

Omnipotence · Jun 28, 2011

Thanks!

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Binomial (1 Viewer)

Omnipotence

Kendrick Lamar

Drongoski

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Drongoski

Well-Known Member

Omnipotence

Kendrick Lamar

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