# Find the 7th term from the end in the expansion

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,

$T_{3}=T_{2+1}$

$={ }^{8} C_{2}\left(2 x^{2}\right)^{8-2}\left(-\frac{3}{2 x}\right)^{2}$

$=\frac{8 \times 7}{2 \times 1}\left(64 x^{12}\right) \frac{9}{4 x^{2}}$

$=4032 x^{10}$