Grazes 2 years ago Find the term in the binomial expansion of (((x^2)/3)-(1/x))^9 that is a constant.

1. Grazes

$(\frac{ x ^{2} }{ 3 } - \frac{ 1 }{ x })^{9}$

2. hewsmike

Nasty one. The binomial expansion$(a+b)^{n}= \sum_{k=0}^{n}\frac{ n! }{k!(n-k)!}a^{n-k}b^{k}$has powers of a decreasing while those of b increase as you go up in the index k. Consider for this problem that $a=x^{2}/3$and $b= -1/x=-x^{-1}$and $n=9$They are effectively asking for the value of k that gives$(a)^{n-k}(b)^{k}=(x^{2}/3)^{9-k}(-x^{-1})^{k}=constant \times x^{0}$and so looking at those powers$2(9-k)+(-k)=0$$18-2k-k=0$$3k=18$$k=3$so what is the multiplier of that power of x?$(1/3)^{9-3}(-1)^{3}\frac{9!}{3!(9-3)!}=(-1)(3^{-6})\frac{(9\times 8 \times 7 \times 6 \times 5\times4 \times 3 \times 2 \times 1)}{(3 \times 2 \times 1)(6 \times 5\times4 \times 3 \times 2 \times 1)}$.... aaargh ......$=-28/3^{5}$I think ! :-)

3. Grazes

I just know that the final answer is 29/8. Thanks!

4. hewsmike

Ah, I've probably fumbled the arithmetic somewhere, but the method is as described. :-)