First slide

Binomial theorem for positive integral Index

Question

The term independent of x in the expansion of $(\frac{1}{60} - \frac{x^{8}}{81}) \cdot {(2 x^{2} - \frac{3}{x^{2}})}^{6}$ is equal to

Moderate

A

-72
B

36
C

-36
D

-108

Solution

Let a binomial ${(2 x^{2} - \frac{3}{x^{2}})}^{6}, it's (r + 1) th term$
$\begin{aligned} = T_{r + 1} =^{6} C_{r} {(2 x^{2})}^{6 - r} {(- \frac{3}{x^{2}})}^{r} \\ =^{6} C_{r} (- 3)^{r} (2)^{6 - r} x^{12 - 2 r - 2 r} - - - (i) \\ =^{6} C_{r} (- 3)^{r} (2)^{6 - r} x^{12 - 4 r} \end{aligned}$
Now, the term independent of x in the expansion of
$(\frac{1}{60} - \frac{x^{8}}{81}) {(2 x^{2} - \frac{3}{x^{2}})}^{6}$ = the term independent of x in
the expansion of $\frac{1}{60} {(2 x^{2} - \frac{3}{x^{2}})}^{6}$
the term independent of x in the
expansion of $- \frac{x^{8}}{81} {(2 x^{2} - \frac{3}{x^{2}})}^{6}$
$= \frac{^{6} C_{3}}{60} (- 3)^{3} (2)^{6 - 3} x^{12 - 4 (3)}$
$+ {(- \frac{1}{81})}^{6} C_{5} (- 3)^{5} (2)^{6 - 5} x^{12 - 4 (5)} x^{8}$
$\begin{aligned} = \frac{1}{3} (- 3)^{3} 2^{3} + \frac{3^{5} \times 2 (6)}{81} \\ = 36 - 72 = - 36 \end{aligned}$

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