1
$\begingroup$

Find the exponent for $x^2$ in the expansion of:

$(\frac{x}{3} - \frac{1}{x^3})^{10}$

What trips me up is that the second term is negative. Though I find using the binomial theorem that the expansion should be a sum of terms on the form:

${ 10 \choose k}(\frac{x}{3})^k(-\frac{1}{x^3})^{10-k} $

But according to the solution, that is wrong. The form is supposed to be:

${ 10 \choose k}(\frac{x}{3})^k(\frac{1}{x^3})^{3(10-k)} $

A solution is then found for $k = 8$ which is simple. But I don't understand why my form is wrong and the solutions form is right. Why is the coefficient for the second factor $3(10-k)$ and where did the negative sign go?

The solution for the problem is supposed to be:

$3^{-8}{ 10 \choose 8 }$

$\endgroup$
  • $\begingroup$ Yeah, where did the negative sign go? $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 18:12
  • $\begingroup$ Your answer looks right to me. What's the source of the solution? (For $k=8$ the minus sign will not matter, but where did the $3$ come from?) $\endgroup$ – Ethan Bolker Jan 15 '17 at 18:16
  • 3
    $\begingroup$ Is there perhaps a typo: we have $\left(\frac{1}{x^3}\right)^{10-k} = \left(\frac{1}{x}\right)^{3(10-k)}$ so if it's supposed to be $1/x$ instead of $1/x^3$ inside the bracket then it becomes correct. $\endgroup$ – Winther Jan 15 '17 at 18:18
  • $\begingroup$ @Winther I think so. And one typo in the "solution" makes a second (the missing minus sign) more likely. $\endgroup$ – Ethan Bolker Jan 15 '17 at 18:23
  • $\begingroup$ Yes @Winther is definitely right, thanks. But it wasn't easy for me to see cause I've spent ages on this problem. $\endgroup$ – Björn Lindqvist Jan 15 '17 at 18:40
0
$\begingroup$

$$\begin{align} \left(\frac x3-\frac 1{x^3}\right)^{10} &=\frac {x^{10}}{3^{10}}\left(1-\frac 3{x^4}\right)^{10}\\ &=\frac {x^{10}}{3^{10}}\sum_{r=0}^{10}\binom {10}r\left(- 3x^{-4}\right)^r \end{align}$$ Putting $r=2$ gives $x^2$ term, hence coefficient of $x^2$ is $$\frac 1{3^{10}} \binom{10}2(-3)^2=\frac {45}{3^8}=\color{red}{\frac 5{3^6}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.