Find the coefficient of in the expansion of

I found and could not work with a fraction....

Basically, I equated the powers of x to zero,

and found

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- Aug 6th 2010, 09:21 PMPunchBinomial Expansion(2)
Find the coefficient of in the expansion of

I found and could not work with a fraction....

Basically, I equated the powers of x to zero,

and found - Aug 6th 2010, 09:55 PMundefined
- Aug 6th 2010, 10:00 PMsa-ri-ga-ma
r cannot be a fraction.

22 - 3r = 7.

Now find r. - Aug 7th 2010, 03:38 AMronaldjames500
you can't get r in fraction, it has to be a natural number.

With that you'll get

=> 22-3r = 7

=> 22-7 = 3r

=> 15 = 3r

=> 5 = r - Aug 7th 2010, 10:39 PMPunch
I realised that I equated it to 0, thinking that it was an independent term...

However, what if the questions ask for an independent term, would it be possible? - Aug 7th 2010, 10:52 PMmathaddict
- Aug 8th 2010, 01:18 AMAlso sprach Zarathustra
If it easy for you, we can look at: (2x^2+1/x)11={1/x^11}{2x^3+1)^11, notice that now we need to find the coefficient x^18 of {2x^3+1)^11...