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

To solve the given problem, we need to find the possible values of \( x \) for which the conditions of the binomial expansion hold.

The expansion is given as \((\sqrt{2^{\log_2(10-3^x)}} + \sqrt[5]{2^{(x-2)\log_2{3}}})^m\). Let \( a = \sqrt{2^{\log_2(10-3^x)}} \) and \( b = \sqrt[5]{2^{(x-2)\log_2{3}}} \). The sixth term in the binomial expansion is given by:

\[\binom{m}{5}a^{m-5}b^5\]

.

This equals 21, thus:

\[\binom{m}{5}(\sqrt{2^{\log_2(10-3^x)}})^{m-5}(\sqrt[5]{2^{(x-2)\log_2{3}}})^5=21\]

.

Redefine: \(a^{m-5} = (2^{\log_2{(10-3^x)}})^{(m-5)/2} = (10-3^x)^{(m-5)/2} \) and \( b^5 = 2^{(x-2)\log_2{3}} \).

Simplifying gives:

\[\binom{m}{5}(10-3^x)^{(m-5)/2}2^{(x-2)\log_2{3}} = 21\]

For the coefficients of the binomial expansion:

\[\binom{m}{1}, \binom{m}{2}, \binom{m}{3},\]

 form an AP. Relations:

\[\binom{m}{2} - \binom{m}{1} = \binom{m}{3} - \binom{m}{2}\]

.

Simplified, this gives:

\[(m-1)= (m-2)(m-3)/2 - (m(m-1)/2)\]

 solving confirms \( m = 5 \).

Substitute \( m=5 \) back to find x:

\[10-3^x=1\]

 giving \( 3^x=9 \) hence \( x=2 \).

Summing the squares of possible x values (here only one), the result is \( 2^2 = 4 \) within the range 4 to 4.

Therefore, the sum of the squares of all possible values of \( x \) is \( 4 \).