The term independent of x in the expansion of160−x881⋅2x2−3x26 is equal to

A
-108
B
-72
C
-36
D
36

Solution:

Given expansion is $(\frac{1}{60} - \frac{x^{8}}{81}) {(2 x^{2} - \frac{3}{x^{2}})}^{6}$

$= \frac{1}{60} {(2 x^{2} - \frac{3}{x^{2}})}^{6} - \frac{x^{8}}{81} {(2 x^{2} - \frac{3}{x^{2}})}^{6}$

So its general term is

$\begin{array}{l} T_{r + 1} = {\frac{1}{60}}^{6} C_{r} {(2 x^{2})}^{6 - r} {(\frac{- 3}{x^{2}})}^{r} - \frac{x^{8}}{81} \cdot^{6} C_{r} {(2 x^{2})}^{6 - r} {(\frac{- 3}{x^{2}})}^{r} \\ = {\frac{1}{60}}^{6} C_{r} (2)^{6 - r} (- 3)^{r} x^{12 - 2 r - 2 r} - {\frac{1}{81}}^{6} C_{r} (2)^{6 - r} (- 3)^{r} x^{12 - 2 r - 2 r + 8} \\ = {\frac{1}{60}}^{6} C_{r} (2)^{6 - r} (- 3)^{r} x^{12 - 4 r} - {\frac{1}{81}}^{6} C_{r} (2)^{6 - r} (- 3)^{r} x^{20 - 4 r} \end{array}$

For this term to be independent of $x,$ put $r = 3$ in

1st part and r = 5 in 2nd part.

From , $(i), T_{r + 1} = \frac{1}{60} \times 20 \times 8 (- 27) + \frac{1}{81} \times 6 \times 2 \times 243$

$= - 72 + 36 = - 36$

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

Solution:

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