Binomial Theorem 1 Question 9
10.
The ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}})^{10}$ is
(2019 Main, 12 Jan I)
(a) $1: 2(6)^{\frac{1}{3}}$
(b) $1: 4(16)^{\frac{1}{3}}$
(c) $4(36)^{\frac{1}{3}}: 1$
(d) $2(36)^{\frac{1}{3}}: 1$
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Answer:
Correct Answer: 10. (c)
Solution:
- Since, $r$ th term from the end in the expansion of a binomial $(x+a)^{n}$ is same as the $(n-r+2)$ th term from the beginning in the expansion of same binomial.
$\therefore$ Required ratio $=\frac{T_{5}}{T_{10-5+2}}=\frac{T_{5}}{T_{7}}=\frac{T_{4+1}}{T_{6+1}}$
$\Rightarrow \frac{T_{5}}{T_{10-5+2}} =\frac{{ }^{10} C_{4}\left(2^{1 / 3}\right)^{10-4} (\frac{1}{2(3)^{1 / 3}})^4}{{ }^{10} C_{6}\left(2^{1 / 3}\right)^{10-6} (\frac{1}{2(3)^{1 / 3}})^6} \quad\left[\because T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r}\right]$
$ =\frac{2^{6 / 3}\left(2(3)^{1 / 3}\right)^{6}}{2^{4 / 3}\left(2(3)^{1 / 3}\right)^{4}} \quad\left[\because{ }^{10} C_{4}={ }^{10} C_{6}\right] $
$ =2^{6 / 3-4 / 3}\left(2(3)^{1 / 3}\right)^{6-4} $
$ =2^{2 / 3} \cdot 2^{2} \cdot 3^{23}=4(6)^{2 / 3}=4(36)^{1 / 3}$
So, the required ratio is $4(36)^{1 / 3}: 1$.
