Understanding the Binomial Expansion and A.P. Properties

This problem combines concepts from the Binomial Theorem, properties of Arithmetic Progressions (A.P.), and logarithmic/exponential functions. The general term in the binomial expansion of $(A+B)^m$ is given by the formula: $T_{r+1} = {^m C_r} A^{m-r} B^r$ where ${^m C_r} = \frac{m!}{r!(m-r)!}$ is the binomial coefficient. For three terms $a, b, c$ to be in an Arithmetic Progression, the middle term is the average of the first and last term, i.e., $2b = a+c$.

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Step 1: Determine the value of 'm' using the A.P. condition

The problem states that the binomial coefficients of the second, third, and fourth terms are in an A.P.

• The coefficient of the second term ($T_2$) is ${^m C_1}$.
• The coefficient of the third term ($T_3$) is ${^m C_2}$.
• The coefficient of the fourth term ($T_4$) is ${^m C_3}$.

Since these three coefficients are in A.P., we can write: $2 \cdot ({^m C_2}) = {^m C_1} + {^m C_3}$

Now, let's expand these binomial coefficients:

• ${^m C_1} = m$
• ${^m C_2} = \frac{m(m-1)}{2 \cdot 1} = \frac{m(m-1)}{2}$
• ${^m C_3} = \frac{m(m-1)(m-2)}{3 \cdot 2 \cdot 1} = \frac{m(m-1)(m-2)}{6}$

Substitute these into the A.P. equation: $2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}$ $m(m-1) = m + \frac{m(m-1)(m-2)}{6}$

Explanation: We simplify the expression to solve for $m$. Since binomial coefficients are defined for $m \ge r$, and we have ${^m C_3}$, $m$ must be at least 3. This means $m \neq 0$. Therefore, we can safely divide both sides by $m$: $(m-1) = 1 + \frac{(m-1)(m-2)}{6}$

Now, multiply the entire equation by 6 to eliminate the denominator: $6(m-1) = 6 + (m-1)(m-2)$ $6m - 6 = 6 + m^2 - 2m - m + 2$ $6m - 6 = m^2 - 3m + 8$

Rearrange the terms to form a standard quadratic equation: $m^2 - 3m - 6m + 8 + 6 = 0$ $m^2 - 9m + 14 = 0$

Factor the quadratic equation: $(m-7)(m-2) = 0$

This gives two possible values for $m$: $m=7$ or $m=2$. Explanation: As established earlier, for ${^m C_3}$ to exist and be a valid term in the expansion (i.e., for the fourth term to exist), $m$ must be greater than or equal to 3. Therefore, $m=2$ is not a valid solution. Thus, the only valid value for $m$ is $m=7$.

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Step 2: Simplify the terms of the binomial expression

The given binomial expression is ${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$. Let's simplify the first term, $A = \sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}}$. Explanation: We use the logarithm property $a^{\log_a b} = b$. $A = \sqrt{10 - {3^x}}$ Condition: For the logarithm to be defined, $10 - 3^x > 0$, which implies $3^x < 10$. We must check this condition for our final values of $x$.

Now, let's simplify the second term, $B = \root 5 \of {{2^{(x - 2){{\log }_2}3}}}$. Explanation: We use the logarithm property $k \log_a b = \log_a b^k$, and then $a^{\log_a b} = b$. $B = \root 5 \of {{2^{{{\log }_2}\left( {{3^{(x - 2)}}} \right)}}}$ $B = \root 5 \of {{3^{(x - 2)}}}$ $B = \left(3^{(x-2)}\right)^{1/5}$

So the binomial expansion is of the form $\left( \sqrt{10 - 3^x} + (3^{x-2})^{1/5} \right)^7$. The problem specifies "in the increasing powers of $2^{(x-2) \log_2 3}$," which corresponds to our simplified $B = (3^{x-2})^{1/5}$. This means $A$ is the first term and $B$ is the second term in the general formula $(A+B)^m$.

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Step 3: Apply the Binomial Theorem for the sixth term

We need to find the sixth term ($T_6$) in the expansion. Using the formula $T_{r+1} = {^m C_r} A^{m-r} B^r$, for $T_6$, we have $r+1 = 6$, which means $r=5$. Substituting $m=7$, $r=5$, $A = \sqrt{10 - 3^x}$, and $B = (3^{x-2})^{1/5}$: $T_6 = {^7 C_5} \left(\sqrt{10 - 3^x}\right)^{7-5} \left((3^{x-2})^{1/5}\right)^5$ $T_6 = {^7 C_5} \left(\sqrt{10 - 3^x}\right)^2 \left(3^{x-2}\right)$

Explanation: We simplify the powers. The square root and the power of 2 cancel out, and the fifth root and power of 5 cancel out. $T_6 = {^7 C_5} (10 - 3^x) (3^{x-2})$

Now, calculate the binomial coefficient ${^7 C_5}$: ${^7 C_5} = {^7 C_{7-5}} = {^7 C_2} = \frac{7 \times 6}{2 \times 1} = 21$

Substitute this value back into the expression for $T_6$: $T_6 = 21 (10 - 3^x) (3^{x-2})$

We are given that the sixth term is $21$. So, we set up the equation: $21 (10 - 3^x) (3^{x-2}) = 21$

Explanation: Since $21 \ne 0$, we can divide both sides by 21 to simplify the equation. $(10 - 3^x) (3^{x-2}) = 1$

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Step 4: Solve the equation for 'x'

We have the equation: $(10 - 3^x) (3^{x-2}) = 1$

Explanation: To solve this, we will rewrite $3^{x-2}$ using the exponent rule $a^{b-c} = a^b \cdot a^{-c}$. $3^{x-2} = 3^x \cdot 3^{-2} = \frac{3^x}{3^2} = \frac{3^x}{9}$

Substitute this back into the equation: $(10 - 3^x) \left(\frac{3^x}{9}\right) = 1$

Now, to make it easier to solve, let $y = 3^x$. $(10 - y) \left(\frac{y}{9}\right) = 1$

Multiply both sides by 9: $y(10 - y) = 9$ $10y - y^2 = 9$

Rearrange the terms to form a quadratic equation in $y$: $y^2 - 10y + 9 = 0$

Factor the quadratic equation: $(y-9)(y-1) = 0$

This gives two possible values for $y$:

• $y = 9$
• $y = 1$

Now, substitute back $y = 3^x$ to find the values of $x$: Case 1: $3^x = 9$ $3^x = 3^2$ $x = 2$

Case 2: $3^x = 1$ $3^x = 3^0$ $x = 0$

Check for validity: Recall the condition from Step 2: $3^x < 10$.

• For $x=2$, $3^2 = 9$. Since $9 < 10$, $x=2$ is a valid solution.
• For $x=0$, $3^0 = 1$. Since $1 < 10$, $x=0$ is a valid solution. Both values of $x$ are possible.

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Step 5: Calculate the sum of the squares of all possible values of 'x'

The possible values for $x$ are $0$ and $2$. Sum of the squares of these values $= 0^2 + 2^2 = 0 + 4 = 4$.

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Tips and Common Mistakes:

• Checking domain constraints: Always verify that your solutions for $x$ satisfy the original domain constraints of the problem, especially for logarithmic and root functions. In this case, $10 - 3^x > 0$ was crucial.
• Validity of 'm': When solving for 'm' (or 'n' in general), ensure the value obtained allows for all binomial coefficients mentioned in the problem to be valid (e.g., $m \ge r$).
• Careful with exponents and logarithms: Mistakes often occur when simplifying terms like $2^{(x-2)\log_2 3}$. Remember the properties $k \log_a b = \log_a b^k$ and $a^{\log_a b} = b$.
• Reading the problem carefully: Note the phrase "increasing powers of..." to correctly identify which term is $A$ and which is $B$ in $(A+B)^m$.

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Summary and Key Takeaway: This problem is an excellent example of how different mathematical concepts (Binomial Theorem, Arithmetic Progressions, and properties of exponents and logarithms) are integrated into a single question. The key to solving such problems lies in systematically breaking down the problem, carefully applying the relevant formulas and properties at each step, and diligently checking for any domain constraints or conditions that might limit the possible solutions. The final answer is obtained by combining the valid solutions derived from each part of the problem.