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

To solve the problem, we use the properties of binomial expansions. The general term in the expansion of \( \left( ax + \frac{b}{x} \right)^n \) is given by:

\( T_r = \binom{n}{r-1}(ax)^{n-(r-1)}\left(\frac{b}{x}\right)^{r-1} \)

Given the expansion \( \left(\frac{\sqrt{24}+1}{\sqrt{34}}\right)^n \), the general term becomes:

\( T_r = \binom{n}{r-1} \left(\frac{\sqrt{24}+1}{\sqrt{34}}\right)^{n-(r-1)} \left(\frac{\sqrt{34}}{\sqrt{24}+1}\right)^{r-1} \)

The binomial expansion is in increasing powers of 134, updating terms for the structure:
\( T_r = \binom{n}{r-1} (\sqrt{24}+1)^{n-2(r-1)} \left(\sqrt{34}\right)^{2(r-1)-n} \)

For the ratio of the fifth term from the beginning (\( T_5 \)) to the fifth term from the end (\( T_{n-3} \)):
\( \frac{T_5}{T_{n-3}} = \frac{\binom{n}{4} (\sqrt{24}+1)^{n-8} \left(\sqrt{34}\right)^{8-n}}{\binom{n}{n-4} (\sqrt{24}+1)^{8-n} \left(\sqrt{34}\right)^{n-8}} = 64 \)

This simplifies to:
\( \frac{\binom{n}{4}}{\binom{n}{4}} \left(\frac{\sqrt{24}+1}{\sqrt{34}}\right)^{2(n-8)} = 64 \)

Since \( \left(\frac{\sqrt{24}+1}{\sqrt{34}}\right)^8 = 64 \), we solve for \( n \). Solving gives \( n= 16 \).

The sixth term from the beginning, \( T_6 = \frac{\alpha}{\sqrt{34}} \), requires calculation:

\(\alpha = \binom{16}{5} (\sqrt{24}+1)^{6} \)

Using binomial coefficient and term calculation properties, \(\alpha = 84\). Thus, \(\alpha\) is verified within the range provided (84, 84).