SATHEE: Binomial Theorem 1 Question 9

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

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

Show Answer

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.

$∴$ Required ratio $= \frac{T_{5}}{T_{10 - 5 + 2}} = \frac{T_{5}}{T_{7}} = \frac{T_{4 + 1}}{T_{6 + 1}}$

$\Rightarrow \frac{T_{5}}{T_{10 - 5 + 2}} = \frac{^{10} C_{4} {(2^{1 / 3})}^{10 - 4} (\frac{1}{2 (3)^{1 / 3}})^{4}}{^{10} C_{6} {(2^{1 / 3})}^{10 - 6} (\frac{1}{2 (3)^{1 / 3}})^{6}} [∵ T_{r + 1} =^{n} C_{r} x^{n - r} a^{r}]$

$= \frac{2^{6 / 3} {(2 (3)^{1 / 3})}^{6}}{2^{4 / 3} {(2 (3)^{1 / 3})}^{4}} [∵^{10} C_{4} =^{10} C_{6}]$

$= 2^{6 / 3 - 4 / 3} {(2 (3)^{1 / 3})}^{6 - 4}$

$= 2^{2 / 3} \cdot 2^{2} \cdot 3^{23} = 4 (6)^{2 / 3} = 4 (36)^{1 / 3}$

So, the required ratio is $4 (36)^{1 / 3} : 1$ .

Binomial Theorem 1 Question 9

10.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Binomial Theorem 1 Question 9

10.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu