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

We have 21/4+131/4nGiven that T5Tn−3=61/41⇒ nC421/4n−43−1/44 nCn−421/443−1/4n−4=61/41⇒21/4n−83−1/48−n=61/4⇒61/4n−8=61/4⇒n−8=1⇒n=9Now, T6=nC521/443−1/45=126×2×3−13−1/4=2523131/4=8434=α34Thus, α=84

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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________

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answer is 84.

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Detailed Solution

We have ${(2^{1 / 4} + \frac{1}{3^{1 / 4}})}^{n}$
Given that $\frac{T_{5}}{T_{n - 3}} = \frac{6^{1 / 4}}{1}$
$\begin{array}{l} \Rightarrow \frac{^{n} C_{4} {(2^{1 / 4})}^{n - 4} {(3^{- 1 / 4})}^{4}}{^{n} C_{n - 4} {(2^{1 / 4})}^{4} {(3^{- 1 / 4})}^{n - 4}} = \frac{6^{1 / 4}}{1} \\ \Rightarrow {(2^{1 / 4})}^{n - 8} {(3^{- 1 / 4})}^{8 - n} = 6^{1 / 4} \\ \Rightarrow {(6^{1 / 4})}^{n - 8} = 6^{1 / 4} \Rightarrow n - 8 = 1 \Rightarrow n = 9 \end{array}$
Now, $T_{6} =^{n} C_{5} {(2^{1 / 4})}^{4} {(3^{- 1 / 4})}^{5}$
$= 126 \times 2 \times (3^{- 1}) (3^{- 1 / 4}) = \frac{252}{3} \frac{1}{3^{1 / 4}} = \frac{84}{\sqrt[4]{3}} = \frac{α}{\sqrt[4]{3}}$
Thus, $α = 84$