Find the 7^{th} term from the end in the expansion of ${(2 x^{2} - \frac{3}{2 x})}^{8}$

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Solution

$7^{t h}$ term from the end $= 3^{r d}$ term from beginning

$T_{N} = T_{r + 1} = (- 1)^{r}^{n} C_{2} x^{[} n - r] y^{r}$

$N = 3, r = 2, n = 8, x = 2 x^{2}, y = \frac{3}{2 x}$

$T_{3} = T_{2 + 1} = (- 1)^{2}^{8} C_{2} (2 x^{2})^{6} {(\frac{3}{2 x})}^{2}$

$\frac{8 \times 7}{2} \times \frac{2^{6} \times 3^{2} \times x^{12}}{2^{2} \times x^{2}} = 8 \times 7 \times 9 \times 8 \times x^{10}$

$= 4032 x^{10}$

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