Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of \(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}\), in the increasing powers of \(\frac{1}{\sqrt[4]{3}}\) be \(\sqrt[4]{6}: 1\). If the sixth term from the beginning is \(\frac{\alpha}{\sqrt[4]{3}}\), then \(\alpha\) is equal to ______.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Binomial Theorem

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of

(42 + \frac{1}{4 3})^{n}

, in the increasing powers of

\frac{1}{4 3}

46 : 1

. If the sixth term from the beginning is

\frac{α}{4 3}

, then

α

is equal to ______.

Answer: 84

The fifth term from the beginning is

T_{5} =^{n} C_{4} (2^{1/4})^{n - 4} (3^{- 1/4})^{4}

and the fifth term from the end is

T_{n - 3} =^{n} C_{n - 4} (2^{1/4})^{4} (3^{- 1/4})^{n - 4}

. Setting their ratio equal to

46

\frac{T _{5}}{T _{n - 3}} = \frac{2 ^{(n - 8) /4} \cdot 3 ^{(n - 8) /4}}{1} = \frac{6 ^{(n - 8) /4}}{1} = 6^{1/4}

This gives

n - 8 = 1

, so

n = 9

.

The sixth term is:

T_{6} =^{9} C_{5} (2^{1/4})^{4} (3^{- 1/4})^{5} = 126 \cdot 2 \cdot 3^{- 5/4} = \frac{252}{3 ^{5/4}} = \frac{252}{3 \cdot 3 ^{1/4}} = \frac{84}{4 3}

Therefore

α = 84

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer