Question

Let the sixth term in the binomial expansion of \(({\sqrt{2}^{log_{2}}(10-3^{x})+\sqrt[5]{2^{(x-2)log_{2}{3}}}})^{m}\), in the increasing powers of \(2^{(x-2)log_{2}3}\), be 21 If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an AP, then the sum of the squares of all possible values of x is

Hint:

In problems involving binomial expansions with parameters, always analyze the given constraints and relationships among coefficients carefully.

Answer 1

Correct Answer: 4

The sixth term in the binomial expansion is given by:
\[T_6 = \binom{m}{5}(10 - 3x)^{m-5} \cdot (3x^{-2})^5 = 21 \quad \text{...(1)}.\]
It is given that:
\[\binom{m}{1}, \binom{m}{2}, \binom{m}{3}\]
are in an arithmetic progression (A.P.).
Step 1: Solve for \(m\) from the A.P. condition
From the property of an A.P.:
\[2 \binom{m}{2} = \binom{m}{1} + \binom{m}{3}.\]
Using the formula for binomial coefficients:
\[2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}.\]
Simplifying:
\[m(m-1) = 2m + \frac{m(m-1)(m-2)}{3}.\]
\[3m(m-1) = 6m + m(m-1)(m-2).\]
\[m(m-1)(m-2) - 3m(m-1) + 6m = 0.\]
\[m(m-1)(m-5) = 0.\]
Thus, \(m = 0, 2, 7\). Excluding \(m = 0\) (trivial case) and \(m = 2\) (does not satisfy the conditions), we get \(m = 7\).
Step 2: Substitute \(m = 7\) into equation (1)
Substituting \(m = 7\) in equation (1):
\[\binom{7}{5} \cdot (10 - 3x)^2 \cdot \frac{(3x^{-2})^5}{9} = 21.\]
\[\binom{7}{5} = \frac{7 \cdot 6}{2} = 21.\]
\[21 \cdot (10 - 3x)^2 \cdot \frac{3^5}{9x^{10}} = 21.\]
\[(10 - 3x)^2 \cdot \frac{3^5}{9x^{10}} = 1.\]
Let \(y = 3x^{-2}\). Then:
\[10y - y^2 = 1.\]
Step 3: Solve for \(y\)
Rewriting:
\[y^2 - 10y + 1 = 0.\]
Using the quadratic formula:
\[y = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2}.\]
\[y = 5 \pm 2\sqrt{6}\]
Step 4: Solve for \(x\)
Since \(y = 3x^{-2}\):
\[3x^{-2} = 5 + 2\sqrt{6} \quad \text{or} \quad 3x^{-2} = 5 - 2\sqrt{6}.\]
For each \(y\), compute \(x\). The sum of the squares of \(x\) values is:
\[\boxed{4}.\]