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

with the LAST n-1 terms so you have

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

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HoldingOn

Active Member
NAH

Jks, this is a pretty cool problem.

You start by applying the symmetrical co-efficient property

with the LAST n-1 terms so you have

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

If we had a even number of terms, would that negate the need to add 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 as the expression would be perfectly symmetrical?
There can never be an even number of terms, only an odd amount.

You're expanding

So expanding generates 3 terms.

expanding 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

The second term looks pretty close to and then a bit of intuition comes into play.

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• HeroWise