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

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Solution

Let T_r₊₁ be the 7th term from the end in the given expression.
Then, we have:
T_r₊₁ = (9 − 7 + 1) = 3rd term from the beginning
Now,
$T_{3} = T_{2 + 1} = {}^{8}C_{2} (2 x^{2})^{8 - 2} {(- \frac{3}{2 x})}^{2} = \frac{8 \times 7}{2 \times 1} (64 x^{12}) \frac{9}{4 x^{2}} = 4032 x^{10}$

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