Binomial Theorem Ques 3
- The term independent of $x$ in the expansion of $\left(\frac{1}{60}-\frac{x^8}{81}\right) \cdot\left(2 x^2-\frac{3}{x^2}\right)^6$ is equal to
(2019 Main, 12 April II)
(a) $-72$
(b) $36$
(c) $-36$
(d) $-108$
Show Answer
Answer:
Correct Answer: 3.(c)
Solution: (c) Key Idea
Use the general term (or $(r+1)$ th term) in the expansion of binomial $(a+b)^n$
i.e. $\quad T_{r+1}={ }^n C_r a^{n-r} b^r$
Let a binomial $\left(2 x^2-\frac{3}{x^2}\right)^6$,
it’s $(r+1)$ th term
$ \begin{aligned} & =T_{r+1}={ }^6 C_r\left(2 x^2\right)^{6-r}\left(-\frac{3}{x^2}\right)^r \\ & ={ }^6 C_r(-3)^r(2)^{6-r} x^{12-2 r-2 r} \\ & ={ }^6 C_r(-3)^r(2)^{6-r} x^{12-4 r} \end{aligned} $
Now, the term independent of $x$ in the expansion of
$\left(\frac{1}{60}-\frac{x^8}{81}\right)\left(2 x^2-\frac{3}{x^2}\right)^6$
$=$ the term independent of $x$ in the expansion of $\frac{1}{60}\left(2 x^2-\frac{3}{x^2}\right)^6+$
the term independent of $x$ in the expansion of $-\frac{x^8}{81}\left(2 x^2-\frac{3}{x^2}\right)^6$
$=\frac{{ }^6 C_3}{60}(-3)^3$ $(2)^{6-3} x^{12-4(3)}$
[put $r=3]$
$\left(-\frac{1}{81}\right){ }^6 C_5(-3)^5$ $(2)^{6-5} x^{12-4(5)} x^8$
[put $r=5]$
$=\frac{1}{3}(-3)^3 2^3+\frac{3^5 \times 2(6)}{81}$ $=36-72=-36$
BETA
