Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of , in the increasing powers of be . If the sixth term from the beginning is , then is equal to

Question

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

Smart Achievers · Accepted Answer

(84)

$We have,$ ${(2^{1 / 4} + \frac{1}{3^{1 / 4}})}^{n}$

$Given that$ $\frac{T_{5}}{T_{n - 3}} = \frac{6^{1 / 4}}{1}$

$\Rightarrow$ $\frac{{}^{n}C_{4} {(2^{1 / 4})}^{n - 4} {(3^{- 1 / 4})}^{4}}{{}^{n}C_{n - 4} {(2^{1 / 4})}^{4} {(3^{- 1 / 4})}^{n - 4}} = \frac{6^{1 / 4}}{1}$

$\Rightarrow$ ${(2^{1 / 4})}^{n - 8} {(3^{- 1 / 4})}^{8 - n} = 6^{1 / 4}$

$\Rightarrow$ ${(6^{1 / 4})}^{n - 8} = 6^{1 / 4} \Rightarrow n - 8 = 1 \Rightarrow n = 9$

$Now,$ $T_{6} = {}^{9}C_{5} {(2^{1 / 4})}^{4} {(3^{- 1 / 4})}^{5}$

$= 126 \times 2 \times (3^{- 1}) (3^{- 1 / 4})$

$= \frac{252}{3} \frac{1}{3^{1 / 4}} = \frac{84}{\sqrt[4]{3}} = \frac{α}{\sqrt[4]{3}}$

$Thus,$ $α = 84$

Solution

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Solution

Q. Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (24+134)n, in the increasing powers of 134 be 64:1. If the sixth term from the beginning is α34, then α is equal to _______.

Ans. (84) We have, (21/4+131/4)n Given that T5Tn-3=61/41 ⇒ C4n(21/4)n-4(3-1/4)4Cn-4n(21/4)4(3-1/4)n-4=61/41 ⇒ (21/4)n-8(3-1/4)8-n=61/4 ⇒ (61/4)n-8=61/4⇒n-8=1⇒n=9 Now, T6=C59(21/4)4(3-1/4)5 =126×2×(3-1)(3-1/4) =2523131/4=8434=α34 Thus, α=84

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