The term independent of x in the expansion of 160−x881⋅2x2−3x26is equal to

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

$= \frac{^{6} C_{3}}{60} (- 3)^{3} (2)^{6 - 3} x^{12 - 4 (3)}$

$+ {(- \frac{1}{81})}^{6} C_{5} (- 3)^{5} (2)^{6 - 5} x^{12 - 4 (5)} x^{8}$

$\begin{aligned} = \frac{1}{3} (- 3)^{3} 2^{3} + \frac{3^{5} \times 2 (6)}{81} \\ = 36 - 72 = - 36 \end{aligned}$

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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