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**Punch** Find the coefficients of $\displaystyle x^7$ in the expansion of $\displaystyle (1-3x^2)(x^2-\frac{2}{x})^7$

My attempt:

to obtain $\displaystyle x^7$, coefficients of $\displaystyle x^5$ and $\displaystyle x^7$ have to be used.

$\displaystyle \color{blue}x^0$ and $\displaystyle \color{blue}x^7$, $\displaystyle \color{blue}x^2$ and $\displaystyle \color{blue}x^5$

$\displaystyle T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r$

the above is fine if you recognise that there is no $\displaystyle \color{blue}x^7$ term when you multiply the binomial expansion by 1

$\displaystyle \color{blue}x^{14-2r}x^{-r}=x^{14-3r}=x^7$ has no natural solution

Then the $\displaystyle \color{blue}x^7$ term is obtained by multiplying the binomial expansion by $\displaystyle \color{blue}-3x^2$

therefore, $\displaystyle 2(7-r)+(-r)=5$

$\displaystyle r=3$

OR $\displaystyle 2(7-r)+(-r)=5$

$\displaystyle r= 2(\frac{1}{3})$ you should have 14-2r-r=14-3r=5, 3r=14-5=9

Puzzled at arriving at $\displaystyle 2\frac{1}{3}$ for $\displaystyle r$.