First slide

Binomial theorem for positive integral Index

Question

If the 6th term from the beginning is equal to the 6th term from the end in the expansion of ${(2^{1 / 5} + \frac{1}{3^{1 / 5}})}^{n},$ then $n$ is equal to

Moderate

A

7
B

9
C

10
D

12

Solution

$6 t h$ term from the beginning
$=^{n} C_{5} {(2^{1 / 5})}^{n - 5} {(\frac{1}{3^{1 / 5}})}^{5} =^{n} C_{5} (\frac{2^{n / 5}}{6})$
Also, $6 t h$ term from the end
$= (n + 2 - 6) = (n - 4) th$ from the beginning
$=^{n} C_{n - 5} {(2^{1 / 5})}^{n - (n - 5)} {(\frac{1}{3^{1 / 5}})}^{n - 5} =^{n} C_{5} (\frac{6}{3^{n / 5}})$
We are given
$^{n} C_{5} (\frac{2^{n / 5}}{6}) =^{n} C_{5} (\frac{6}{3^{n / 5}}) \Rightarrow 6^{n / 5} = 6^{2} \Rightarrow n = 10$

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