Thread: Binomial 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

Originally Posted by Punch

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

Sorry but what does r represent? It might help you to rewrite as .

r cannot be a fraction.



22 - 3r = 7.

Now find r.

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

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?

Originally Posted by Punch

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?

yes in that case.

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...