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Question

Mathematics Question on Binomial theorem

The term independent of x in the expansion of (160x881).(2x23x2)6\bigg(\frac{1}{60} - \frac{x^8}{81}\bigg). \bigg(2x^2 - \frac{3}{x^2}\bigg)^6 is equal to:

A

36

B

-108

C

-72

D

-36

Answer

-36

Explanation

Solution

160(2x23x2)6181.x8(2x23x2)6\frac{1}{60} \bigg(2x^2 - \frac{3}{x^2}\bigg)^6 \, \, - \frac{1}{81}. x^8 \bigg(2x^2 - \frac{3}{x^2}\bigg)^6 its general term 1606Cr26r(3)rx12r1816Cr26r(3)r12204r\frac{1}{60} ^{6}C_{r} 2^{6-r} (-3)^r \, x^{12-r} \, - \frac{1}{81} ^6C _{r} 2^{6-r} (-3)^r 12^{20-4r} for term independent of x,rx, r for IstI^{st} expression is 33 and rr for second expression is 55 \therefore term independent of x=36x = - 36