1. ## Binomial Expansion(2)

Find the coefficient of $x^7$ in the expansion of $(2x^2+\frac{1}{x})^{11}$

I found $r=7\frac{1}{3}$ and could not work with a fraction....

Basically, I equated the powers of x to zero,

$2(11-r)-r=0$

and found $r=7\frac{1}{3}$

2. Originally Posted by Punch
Find the coefficient of $x^7$ in the expansion of $(2x^2+\frac{1}{x})^{11}$

I found $r=7\frac{1}{3}$ and could not work with a fraction....

Basically, I equated the powers of x to zero,

$2(11-r)-r=0$

and found $r=7\frac{1}{3}$
Sorry but what does r represent? It might help you to rewrite $\frac{1}{x}$ as $x^{-1}$.

3. r cannot be a fraction.

$(x^2)^{(11-r)}(x)^{-1} = x^7$

22 - 3r = 7.

Now find r.

4. 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

5. 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?

6. 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.

7. 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...