First slide

Binomial theorem for positive integral Index

Question

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${(2^{\frac{1}{3}} + \frac{1}{2 (3)^{\frac{1}{3}}})}^{10} is$

Moderate

A

$1 : 2 (6)^{\frac{1}{3}}$
B

$1 : 4 (16)^{\frac{1}{3}}$
C

$4 (36)^{\frac{1}{3}} : 1$
D

$2 (36)^{\frac{1}{3}} : 1$

Solution

Since, rth term from the end in the expansion of a binomial (r + $α$ )ⁿ is same as the (n - r +2)th term from the beginning in the expansion of same binomial.
$\begin{aligned} ∴ Required ratio = \frac{T_{5}}{T_{10 - 5 + 2}} \\ = \frac{T_{5}}{T_{7}} = \frac{T_{4 + 1}}{T_{6 + 1}} \end{aligned}$
$\Rightarrow \frac{T_{5}}{T_{10 - 5 + 2}} = \frac{^{10} C_{4} {(2^{1 / 3})}^{10 - 4} {(\frac{1}{2 (3)^{1 / 3}})}^{4}}{^{10} C_{6} {(2^{1 / 3})}^{10 - 6} {(\frac{1}{2 (3)^{1 / 3}})}^{6}}$
$\begin{array}{r} [∵ T_{r + 1} =^{n} C_{r} x^{n - r} a^{r}] \\ [∵^{10} C_{4} =^{10} C_{6}] \end{array}$
$\begin{aligned} = \frac{2^{6 / 3} {(2 (3)^{1 / 3})}^{6}}{2^{4 / 3} {(2 (3)^{1 / 3})}^{4}} \\ = 2^{6 / 3 - 4 / 3} {(2 (3)^{1 / 3})}^{6 - 4} \\ = 2^{23} \cdot 2^{2} \cdot 3^{23} \\ = 4 (6)^{23} \\ = 4 (36)^{1 / 3} \end{aligned}$
So, the required ratio is. $4 (36)^{1 / 3} : 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App