Question

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (\(\frac{\sqrt{24}+1}{\sqrt{34}}\))n, in the increasing powers of 134 be 64:1.If the sixth term from the beginning is \(\frac{\alpha}{\sqrt{34}}\), then α is equal to ___________.

Answer 1

Correct Answer: 84

Fifth term from the beginning
=ⁿC₄(\(\frac{21}{4}\))_n−4(3−\(\frac{1}{4}\))₄
Fifth term from end = (n – 5 + 1)^th term from begin
=ⁿC_n−4(\(\frac{21}{4}\))₃(\(\frac{3-1}{4}\))_n−4
Given
ⁿC₄2_n−44⋅3−\(^{\frac{1}{n}}\)C_n−\(\frac{324}{4}\)⋅3−(\(\frac{n-4}{4}\))=\(\frac{64}{4}\)
⇒\(\frac{6n-8}{4}\)=\(\frac{61}{4}\)
⇒\(\frac{n-8}{4}\)=\(\frac{1}{4}\) ⇒n=9
T₆=T₅₊₁=⁹C₅(\(\frac{21}{4}\))4(3−\(\frac{1}{4}\))₅
=⁹C₅⋅\(\frac{231}{4.3}\)=\(\frac{8431}{4}\)=\(\frac{\alpha31}{4}\)
⇒ α = 84.

Answer 2

Binomial Expansion Term Calculation 

Let T_r denote the r^th term from the beginning in the binomial expansion of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$. Then

$T_{r+1} = {}^nC_r (\sqrt[4]{2})^{n-r} (\frac{1}{\sqrt[4]{3}})^r = {}^nC_r (2)^{\frac{n-r}{4}} (3)^{-\frac{r}{4}}$.

The fifth term from the beginning is T₅, so r = 4:
$T_5 = {}^nC_4 (2)^{\frac{n-4}{4}} (3)^{-1}$.

The fifth term from the end is T'_n-4 = T_n-3, so r = n − 4:
$T_{n-3} = {}^nC_{n-4} (2)^{\frac{n-(n-4)}{4}} (3)^{-\frac{n-4}{4}} = {}^nC_4 (2)^{\frac{1}{}} (3)^{\frac{4-n}{4}}$.

We are given that $\frac{T_5}{T_{n-3}} = \sqrt{6}$. Therefore,

$\frac{{}^nC_4 (2)^{\frac{n-4}{4}} (3)^{-1}}{{}^nC_4 (2)^{\frac{1}{}} (3)^{\frac{4-n}{4}}} = \sqrt{6} \Rightarrow \frac{(2)^{\frac{n-4}{4}} 3^{-1}}{(2)^{\frac{1}{}} 3^{\frac{4-n}{4}}} = 2^{\frac{n-4}{4} - \frac{1}{4}} 3^{-1 - \frac{4-n}{4}} = 2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}} = (2 \cdot 3)^{\frac{n-8}{4}} = 6^{\frac{n-8}{4}} = \sqrt{6}$.

Since $\sqrt{6} = 6^{1/2}$, we have:
$6^{\frac{n-8}{4}} = 6^{\frac{1}{2}} \Rightarrow \frac{n - 8}{4} = \frac{1}{2} \Rightarrow n - 8 = 2 \Rightarrow n = 10$.

The third term from the beginning is T₃, so r = 2:
$T_3 = {}^{10}C_2 (2)^{\frac{10-2}{4}} (3)^{-\frac{2}{4}} = 45 \cdot 2^2 \cdot 3^{-1/2} = 45 \cdot 4 \cdot \frac{1}{\sqrt{3}} = \frac{180}{\sqrt{3}} = \frac{180\sqrt{3}}{3} = 60\sqrt{3}$.