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