# Binomial (1 Viewer)

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#### integral95

##### Well-Known Member
NAH

Jks, this is a pretty cool problem.

You start by applying the symmetrical co-efficient property

$\binom{n}{r} = \binom{n}{n-r}$ with the LAST n-1 terms so you have

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

Etc etc etc..... and you eventually get on the LHS

$2 [ \binom{2n}{0} + \binom{2n}{1} + \dots + \binom{2n}{n-2} + \binom{2n}{n-1} ] + \binom{2n}{n} = 4^n \\ \\ Add \binom{2n}{n} to both sides to get \\ \\ 2(\binom{2n}{0} + \binom{2n}{1} + \dots + \binom{2n}{n-2} + \binom{2n}{n-1} + \binom{2n}{n}) = 4^n + \binom{2n}{n}$
$Then divide both sides by 2 and expand the RHS binomial to get your result.$

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#### HoldingOn

##### Active Member
NAH

Jks, this is a pretty cool problem.

You start by applying the symmetrical co-efficient property

$\binom{n}{r} = \binom{n}{n-r}$ with the LAST n-1 terms so you have

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

Etc etc etc..... and you eventually get on the LHS

$2 [ \binom{2n}{0} + \binom{2n}{1} + \dots + \binom{2n}{n-2} + \binom{2n}{n-1} ] + \binom{2n}{n} = 4^n \\ \\ Add \binom{2n}{n} to both sides to get \\ \\ 2(\binom{2n}{0} + \binom{2n}{1} + \dots + \binom{2n}{n-2} + \binom{2n}{n-1} + \binom{2n}{n}) = 4^n + \binom{2n}{n}$
$Then divide both sides by 2 and expand the RHS binomial to get your result.$
If we had a even number of terms, would that negate the need to add $\binom{2n}{n}$ as the expression would be perfectly symmetrical?

#### integral95

##### Well-Known Member
If we had a even number of terms, would that negate the need to add $\binom{2n}{n}$ as the expression would be perfectly symmetrical?
There can never be an even number of terms, only an odd amount.

You're expanding $(1+x)^{even}$

So expanding $(1+x)^2$ generates 3 terms.

expanding $(1+x)^4$ generates 5 terms etc...

You would get a pretty different scenario if you do get an even number of terms.

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#### integral95

##### Well-Known Member
Im just curios integral howd u come to that conclusion?
If you compared the answers in i) and ii)
The last n terms on the left side of the equation are gone, the right side looks something pretty close to something being divided by 2 since

$4^n/2 = 2^{2n-1}$

The second term looks pretty close to $\binom{2n}{n}$ and then a bit of intuition comes into play.

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