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Step-by-Step Solution
Step 1: Express the Binomial Expansion and the Given Conditions
The given expression is
\Bigl(\sqrt{2^{\log_{2}(10 - 3^x)}} + \sqrt[5]{2^{(x-2)\log_{2}3}}\Bigr)^{m}.
We are told that:
The sixth term (when written in ascending powers of 2^{(x-2)\log_{2}3} ) is 21.
The binomial coefficients of the second, third, and fourth terms are the first, third, and fifth terms of an arithmetic progression (A.P.).
Step 2: Relate the Binomial Coefficients to an Arithmetic Progression
In the expansion of (a + b)^m , the second, third, and fourth terms involve the binomial coefficients
^mC_1 , ^mC_2 , and ^mC_3 , respectively. According to the problem, these are the 1st, 3rd, and 5th terms of an A.P. If a is the first term of that A.P. and d is the common difference, then:
a = {}^mC_1,
a + 2d = {}^mC_2,
a + 4d = {}^mC_3.
By analyzing these conditions and using known properties of binomial coefficients, one finds that
m = 7 satisfies the requirement that the second, third, and fourth termsβ binomial coefficients form the 1st, 3rd, and 5th terms of an A.P.
Step 3: Identify the Sixth Term and Set It Equal to 21
With m = 7 , the general term of (a + b)^7 is
T_{r+1} = {}^7C_r \, a^{7-r} \, b^r.
The sixth term corresponds to r = 5 , so
T_6 = {}^7C_5 \, \bigl(\sqrt{2^{\log_2(10 - 3^x)}}\bigr)^{7-5} \,\bigl(\sqrt[5]{2^{(x-2)\log_2 3}}\bigr)^{5}.
Simplifications (using properties of exponents and roots) lead to the given statement that this sixth term equals 21:
T_6 = {}^7C_5 \,(10 - 3^x)^{\frac{2}{2}} \,3^{x-2} = 21.
Step 4: Substitute the Binomial Coefficient and Simplify
We know that {}^7C_5 = 21 . Thus,
21 \,(10 - 3^x)^{1} \,3^{x-2} = 21.
Simplifying,
(10 - 3^x)\,3^{x-2} = 1.
Step 5: Rewrite and Form a Quadratic in 3^x
Rewrite 3^{x-2} as \frac{3^x}{3^2} = \frac{3^x}{9} . Therefore, the equation becomes:
(10 - 3^x) \,\frac{3^x}{9} = 1
\quad \Longrightarrow \quad
(10 - 3^x)\,3^x = 9.
Expanding:
10\,3^x - (3^x)^2 = 9
\quad \Longrightarrow \quad
(3^x)^2 - 10\,3^x + 9 = 0.
Step 6: Solve the Quadratic Equation for 3^x
The quadratic equation
(3^x)^2 - 10\,3^x + 9 = 0
factors as:
3^x = 9 \quad \text{or} \quad 3^x = 1.
Hence:
If 3^x = 9 \implies 3^2 = 9 \implies x = 2.
If 3^x = 1 \implies 3^0 = 1 \implies x = 0.
Step 7: Compute the Sum of Squares of the Possible x Values
The two solutions for x are 2 and 0 . The problem asks for the sum of the squares of these values:
0^2 + 2^2 = 0 + 4 = 4.
Final Answer: 4