Detailed Solution: Finding the Relation Between 'a' and 'b' Using Binomial Expansions

1. Introduction to the Binomial Theorem

The core concept for solving this problem is the Binomial Theorem, which provides a formula for expanding expressions of the form $(X+Y)^n$. The general term (or $(r+1)^{th}$ term) in the expansion of $(X+Y)^n$ is given by: $T_{r+1} = {^nC_r X^{n-r} Y^r}$ where ${^nC_r = n!{r!(n-r)!}}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.

To find the coefficient of a specific power of $x$ in an expansion, we first write out the general term, simplify it to gather all powers of $x$, and then equate the exponent of $x$ to the desired power to find the value of $r$. Once $r$ is known, we substitute it back into the general term (excluding the $x$ part) to get the coefficient.

2. Coefficient of $x^7$ in ${[ {a{x^2} + ( {{1 {bx}}} )} ]^{11}}$

Step 1: Write the General Term

For the expression ${[ {a{x^2} + ( {{1 {bx}}} )} ]^{11}}$, we identify:

• $X = a{x^2}$
• $Y = {1 {bx}}$
• $n = 11$

Using the general term formula $T_{r+1} = {^nC_r X^{n-r} Y^r}$: $T_{r+1} = {^{11}C_r (a{x^2})^{11-r} ( {1 {bx}} )^r}$

Now, we simplify this expression by separating the constants, powers of $x$, and powers of $b$: $T_{r+1} = {^{11}C_r a^{11-r} (x^2)^{11-r} (b^{-1}x^{-1})^r}$ $T_{r+1} = {^{11}C_r a^{11-r} x^{2(11-r)} b^{-r} x^{-r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} x^{22-2r-r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} x^{22-3r}}$

Explanation: We distribute the exponents to each factor within $X$ and $Y$, and then combine the terms involving $x$ by adding their exponents. This step is crucial for isolating the power of $x$.

Step 2: Find the value of $r$ for $x^7$

We want the coefficient of $x^7$. Therefore, we equate the exponent of $x$ from the general term to 7: $22 - 3r = 7$ Subtract 22 from both sides: $-3r = 7 - 22$ $-3r = -15$ Divide by -3: $r = 5$

Explanation: Finding the value of $r$ tells us which specific term in the binomial expansion will contain the desired power of $x$.

Step 3: Calculate the Coefficient of $x^7$

Substitute $r = 5$ back into the coefficient part of the general term (i.e., everything except $x^{22-3r}$): Coefficient of $x^7 = {^{11}C_5 a^{11-5} b^{-5}}$ $= {^{11}C_5 a^6 b^{-5}}$ $= {^{11}C_5 a^6{b^5}}$ Let's call this Equation (1): ${Coefficient of x^7 = {^{11}C_5 a^6{b^5}}}$

3. Coefficient of $x^{-7}$ in ${[ {ax - ( {{1 {b{x^2}}}} )} ]^{11}}$

Step 1: Write the General Term

For the expression ${[ {ax - ( {{1 {b{x^2}}}} )} ]^{11}}$, we identify:

• $X = ax$
• $Y = -{1 {b{x^2}}}$ (Note the negative sign!)
• $n = 11$

Using the general term formula $T_{r+1} = {^nC_r X^{n-r} Y^r}$: $T_{r+1} = {^{11}C_r (ax)^{11-r} ( -{1 {b{x^2}}} )^r}$

Now, we simplify this expression, being very careful with the negative sign: $T_{r+1} = {^{11}C_r a^{11-r} x^{11-r} (-1)^r (b^{-1}x^{-2})^r}$ $T_{r+1} = {^{11}C_r a^{11-r} x^{11-r} (-1)^r b^{-r} x^{-2r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} (-1)^r x^{11-r-2r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} (-1)^r x^{11-3r}}$

Explanation: It's crucial to include the negative sign in the $Y$ term when expanding. The $(-1)^r$ factor accounts for the alternating signs in the expansion.

Step 2: Find the value of $r$ for $x^{-7}$

We want the coefficient of $x^{-7}$. Therefore, we equate the exponent of $x$ from the general term to -7: $11 - 3r = -7$ Subtract 11 from both sides: $-3r = -7 - 11$ $-3r = -18$ Divide by -3: $r = 6$

Step 3: Calculate the Coefficient of $x^{-7}$

Substitute $r = 6$ back into the coefficient part of the general term: Coefficient of $x^{-7} = {^{11}C_6 a^{11-6} b^{-6} (-1)^6}$ $= {^{11}C_6 a^5 b^{-6} (1)}$ (Since $(-1)^6 = 1$) $= {^{11}C_6 a^5{b^6}}$ Let's call this Equation (2): ${Coefficient of x^{-7} = {^{11}C_6 a^5{b^6}}}$

4. Equating the Coefficients and Solving for the Relation

The problem states that the coefficient of $x^7$ equals the coefficient of $x^{-7}$. Therefore, we set Equation (1) equal to Equation (2): ${^{11}C_5 a^6{b^5} = {^{11}C_6 a^5{b^6}}}$

Key Identity: Recall the property of binomial coefficients: ${^nC_r = {^nC_{n-r}}}$. Here, ${^{11}C_5 = {^{11}C_{11-5}} = {^{11}C_6}}$.

Using this identity, we can simplify the equation: ${a^6{b^5} = a^5{b^6}}$

Now, we solve for the relationship between $a$ and $b$. Multiply both sides by $b^6$: ${a^6{b^5} b^6 = a^5}$ $a^6 b = a^5$

Assuming $a 0$ (as it's a coefficient, typically non-zero for the term to exist), we can divide both sides by $a^5$: ${a^6 b{a^5} = a^5{a^5}}$ $ab = 1$

Explanation: We used the symmetry property of binomial coefficients to simplify the equation significantly. Then, algebraic manipulation (multiplying by common denominators, dividing by common factors) led us to the final relationship.

5. Tips for Students & Common Mistakes to Avoid

• Sign Errors: Be extremely careful with negative signs, especially when expanding terms like $(X-Y)^n$. The $(-1)^r$ factor is critical.
• Exponent Rules: Double-check your exponent manipulations, especially when combining terms like $(x^m)^p = x^{mp}$ and $x^m x^p = x^{m+p}$.
• Binomial Coefficient Identity: Remember the identity ${^nC_r = {^nC_{n-r}}}$. It can often simplify calculations.
• Identifying X and Y: Clearly identify the base terms $X$ and $Y$ in $(X+Y)^n$ before applying the general term formula.
• Assuming Non-Zero Values: When dividing by variables like $a$ or $b$, it's usually implied in such problems that these coefficients are non-zero.

6. Summary and Key Takeaway

This problem demonstrates the application of the Binomial Theorem to find specific coefficients in an expansion. The key steps involve:

1. Writing the general term for each expansion.
2. Carefully collecting powers of $x$ to determine the exponent.
3. Solving for $r$ by equating the exponent to the desired power.
4. Substituting $r$ back into the coefficient part of the general term.
5. Equating the derived coefficients and simplifying the resulting algebraic expression using binomial identities to find the final relationship.

The final relation between $a$ and $b$ is $ab = 1$. This corresponds to option (D). (Self-correction: The provided correct answer was (A) $a-b=1$. Let me re-check my calculations carefully. The original solution also arrived at $ab=1$. I should re-verify the question's provided correct answer. Based on my calculation and the original solution's calculation, D seems to be the correct option, not A. I will proceed with my derived answer, as the calculation is consistent.)

Let's re-verify the original solution steps to see if there's any discrepancy. Original solution: Coefficient of x^7 = ${}^{11}{C_5}{( a )^6}{( b )^{ - 5}}$ (Matches my calculation) Coefficient of x^-7 = ${}^{11}{C_6}{( a )^5}{( { - 1} )^6}{( b )^{ - 6}}$ (Matches my calculation, and $(-1)^6 = 1$) Equating them: ${}^{11}{C_5}{( a )^6}{( b )^{ - 5}}$ = ${}^{11}{C_6}{( a )^5}{( { - 1} )^6}{( b )^{ - 6}}$ Since ${}^{11}{C_5} = {^{11}C_6}$ $a^6 b^{-5} = a^5 b^{-6}$ Divide by $a^5$ (assuming $a 0$): $a b^{-5} = b^{-6}$ Multiply by $b^6$: $a b^{-5} b^6 = 1$ $ab = 1$

Both my derivation and the original solution arrive at $ab=1$. If the provided correct answer is A ($a-b=1$), then there might be an error in the question or the provided correct answer, as the derivation consistently leads to $ab=1$. I will assume my derivation is correct and that the intended answer is D. If this were an actual test, I would flag the question. Given the task is to rewrite the solution, and the solution provided also gets $ab=1$, I will stick to the derived result.```latex

Detailed Solution: Finding the Relation Between 'a' and 'b' Using Binomial Expansions

1. Introduction to the Binomial Theorem

The core concept for solving this problem is the Binomial Theorem, which provides a formula for expanding expressions of the form $(X+Y)^n$. The general term (or $(r+1)^{th}$ term) in the expansion of $(X+Y)^n$ is given by: $T_{r+1} = {^nC_r X^{n-r} Y^r}$ where ${^nC_r = n!{r!(n-r)!}}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.

To find the coefficient of a specific power of $x$ in an expansion, we first write out the general term, simplify it to gather all powers of $x$, and then equate the exponent of $x$ to the desired power to find the value of $r$. Once $r$ is known, we substitute it back into the general term (excluding the $x$ part) to get the coefficient.

**2. Coefficient of $x^7$ in {[ {a{x^2 + ( {{1 {bx**}} )} ]^{11}}}}

Step 1: Write the General Term

For the expression ${[ {a{x^2} + ( {{1 {bx}}} )} ]^{11}}$, we identify:

Using the general term formula $T_{r+1} = {^nC_r X^{n-r} Y^r}$: $T_{r+1} = {^{11}C_r (a{x^2})^{11-r} ( {1 {bx}} )^r}$ Now, we simplify this expression by separating the constants, powers of $x$, and powers of $b$: $T_{r+1} = {^{11}C_r a^{11-r} (x^2)^{11-r} (b^{-1}x^{-1})^r}$ $T_{r+1} = {^{11}C_r a^{11-r} x^{2(11-r)} b^{-r} x^{-r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} x^{22-2r-r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} x^{22-3r}}$ Explanation: We distribute the exponents to each factor within $X$ and $Y$, and then combine the terms involving $x$ by adding their exponents. This step is crucial for isolating the power of $x$.

Step 2: Find the value of $r$ for $x^7$

We want the coefficient of $x^7$. Therefore, we equate the exponent of $x$ from the general term to 7: $22 - 3r = 7$ Subtract 22 from both sides: $-3r = 7 - 22$ $-3r = -15$ Divide by -3: $r = 5$ Explanation: Finding the value of $r$ tells us which specific term in the binomial expansion will contain the desired power of $x$.

Step 3: Calculate the Coefficient of $x^7$

Substitute $r = 5$ back into the coefficient part of the general term (i.e., everything except $x^{22-3r}$): $Coefficient of x^7 = {^{11}C_5 a^{11-5} b^{-5}}$ $= {^{11}C_5 a^6 b^{-5}}$ $= {^{11}C_5 a^6{b^5}}$ Let's call this Equation (1): $Coefficient of x^7 = {^{11}C_5 a^6{b^5}} (1)$

**3. Coefficient of x^{-7 in {[ {ax - ( {{1 {b{x^2**}}} )} ]^{11}}}}

Step 1: Write the General Term

For the expression ${[ {ax - ( {{1 {b{x^2}}}} )} ]^{11}}$, we identify:

Using the general term formula $T_{r+1} = {^nC_r X^{n-r} Y^r}$: $T_{r+1} = {^{11}C_r (ax)^{11-r} ( -{1 {b{x^2}}} )^r}$ Now, we simplify this expression, being very careful with the negative sign: $T_{r+1} = {^{11}C_r a^{11-r} x^{11-r} (-1)^r (b^{-1}x^{-2})^r}$ $T_{r+1} = {^{11}C_r a^{11-r} x^{11-r} (-1)^r b^{-r} x^{-2r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} (-1)^r x^{11-r-2r}}$ $T_{r+1} = {^{11}C_r a^{11-r} b^{-r} (-1)^r x^{11-3r}}$ Explanation: It's crucial to include the negative sign in the $Y$ term when expanding. The $(-1)^r$ factor accounts for the alternating signs in the expansion.

Step 2: Find the value of $r$ for x^{-7}

We want the coefficient of $x^{-7}$. Therefore, we equate the exponent of $x$ from the general term to -7: $11 - 3r = -7$ Subtract 11 from both sides: $-3r = -7 - 11$ $-3r = -18$ Divide by -3: $r = 6$

Step 3: Calculate the Coefficient of x^{-7}

Substitute $r = 6$ back into the coefficient part of the general term: $Coefficient of x^{-7} = {^{11}C_6 a^{11-6} b^{-6} (-1)^6}$ $= {^{11}C_6 a^5 b^{-6} (1)}$ (Since $(-1)^6 = 1$) $= {^{11}C_6 a^5{b^6}}$ Let's call this Equation (2): $Coefficient of x^{-7} = {^{11}C_6 a^5{b^6}} (2)$

4. Equating the Coefficients and Solving for the Relation

The problem states that the coefficient of $x^7$ equals the coefficient of $x^{-7}$. Therefore, we set Equation (1) equal to Equation (2): ${^{11}C_5 a^6{b^5} = {^{11}C_6 a^5{b^6}}}$ Key Identity: Recall the property of binomial coefficients: ${^nC_r = {^nC_{n-r}}}$. Here, ${^{11}C_5 = {^{11}C_{11-5}} = {^{11}C_6}}$. Using this identity, we can simplify the equation by cancelling the binomial coefficients from both sides: ${a^6{b^5} = a^5{b^6}}$ Now, we solve for the relationship between $a$ and $b$. Multiply both sides by $b^6$: ${a^6{b^5} b^6 = a^5}$ $a^6 b = a^5$ Assuming $a 0$ (as it's a coefficient, typically non-zero for the term to exist), we can divide both sides by $a^5$: ${a^6 b{a^5} = a^5{a^5}}$ $ab = 1$ Explanation: We used the symmetry property of binomial coefficients to simplify the equation significantly. Then, algebraic manipulation (multiplying by common denominators, dividing by common factors) led us to the final relationship.

5. Tips for Students & Common Mistakes to Avoid

6. Summary and Key Takeaway

This problem demonstrates the application of the Binomial Theorem to find specific coefficients in an expansion. The key steps involve:

The final relation between $a$ and $b$ is $ab = 1$. This corresponds to option (D).