# Binomial Expansion(2)

• Aug 6th 2010, 08:21 PM
Punch
Binomial Expansion(2)
Find the coefficient of $\displaystyle x^7$ in the expansion of $\displaystyle (2x^2+\frac{1}{x})^{11}$

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

Basically, I equated the powers of x to zero,

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

and found $\displaystyle r=7\frac{1}{3}$
• Aug 6th 2010, 08:55 PM
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Quote:

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

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

Basically, I equated the powers of x to zero,

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

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

Sorry but what does r represent? It might help you to rewrite $\displaystyle \frac{1}{x}$ as $\displaystyle x^{-1}$.
• Aug 6th 2010, 09:00 PM
sa-ri-ga-ma
r cannot be a fraction.

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

22 - 3r = 7.

Now find r.
• Aug 7th 2010, 02:38 AM
ronaldjames500
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, 09:39 PM
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?
• Aug 7th 2010, 09:52 PM