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Binomial (1 Viewer)

integral95

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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

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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

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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

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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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