Find the 5th term from the end in the expansion of $(3 x - \frac{1}{x^{2}})$

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Solution

Fifth term from the end is $(11 - 5 + 1) = 7^{t h}$ term from beginning $T_{7} = T_{6 + 1} = (- 1)^{r}^{n} C_{2} x^{n - r} y^{r}$

$= (- 1)^{6}^{10} C_{6} (3 x)^{4} {(\frac{1}{x^{2}})}^{6}$

$=^{10} C_{6} \times 3^{4} \times \frac{x^{4}}{x^{12}} = \frac{210 \times 81}{x^{8}} = \frac{17010}{x^{8}}$

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