Question

If the coefficients of $x^{10}$ and $x^{11}$ in the expansion of $(1 + \alpha x + \beta x^2)(1+x)^{11}$ are 396 and 144 respectively, then $\alpha^2 + \beta^2 =$

• A. \( 10 \)
• B. \( 13 \)
• C. \( 25 \)
• D. \( 20 \)

Hint:

When finding the coefficient of \( x^k \) in the product of two polynomials, say \( P(x) \cdot Q(x) \), identify all pairs of terms \( (a_i x^i) \) from \( P(x) \) and \( (b_j x^j) \) from \( Q(x) \) such that \( i+j=k \). The coefficient of \( x^k \) will be the sum of all such products \( a_i b_j \). For more than two polynomials, extend this idea to triplets or higher order combinations. In this problem, treat \( (1 + \alpha x + \beta x^2) \) as \( P(x) \) and \( (1+x)^{11} \) as \( Q(x) \). The term \( 1 \) from \( P(x) \) combines with \( x^{10} \) from \( Q(x) \), \( \alpha x \) from \( P(x) \) combines with \( x^9 \) from \( Q(x) \), and \( \beta x^2 \) from \( P(x) \) combines with \( x^8 \) from \( Q(x) \) to form the \( x^{10} \) term.

Answer 1

The Correct Option is B

Let the given expression be \( E = (1 + \alpha x + \beta x^2)(1+x)^{11} \). We know that the general term in the binomial expansion of \( (1+x)^n \) is \( T_{k+1} = \binom{n}{k}x^k \). So, for \( (1+x)^{11} \), the general term is \( \binom{11}{k}x^k \). The expansion of \( E \) can be written as: \[ E = 1 \cdot (1+x)^{11} + \alpha x \cdot (1+x)^{11} + \beta x^2 \cdot (1+x)^{11} \] Let's find the coefficient of \( x^{10} \): The terms that contribute to \( x^{10} \) are: % Option (A) From \( 1 \cdot (1+x)^{11} \): The term with \( x^{10} \) is \( 1 \cdot \binom{11}{10}x^{10} \). % Option (B) From \( \alpha x \cdot (1+x)^{11} \): The term with \( x^{10} \) is \( \alpha x \cdot \binom{11}{9}x^9 = \alpha \binom{11}{9}x^{10} \). % Option (C) From \( \beta x^2 \cdot (1+x)^{11} \): The term with \( x^{10} \) is \( \beta x^2 \cdot \binom{11}{8}x^8 = \beta \binom{11}{8}x^{10} \). The coefficient of \( x^{10} \) is: \[ C_{10} = \binom{11}{10} + \alpha \binom{11}{9} + \beta \binom{11}{8} \] We know the binomial coefficients: \[ \binom{11}{10} = \binom{11}{1} = 11, \binom{11}{9} = \binom{11}{2} = \frac{11 \times 10}{2} = 55, \binom{11}{8} = \binom{11}{3} = \frac{11 \times 10 \times 9}{6} = 165 \] Given \( C_{10} = 396 \): \[ 11 + 55\alpha + 165\beta = 396 \] Divide by 11: \[ 1 + 5\alpha + 15\beta = 36 \Rightarrow 5\alpha + 15\beta = 35 \Rightarrow \alpha + 3\beta = 7 \text{(Equation 1)} \] Now, let's find the coefficient of \( x^{11} \): The terms that contribute to \( x^{11} \) are: % Option (A) From \( 1 \cdot (1+x)^{11} \): The term with \( x^{11} \) is \( \binom{11}{11}x^{11} \). % Option (B) From \( \alpha x \cdot (1+x)^{11} \): The term with \( x^{11} \) is \( \alpha \binom{11}{10}x^{11} \). % Option (C) From \( \beta x^2 \cdot (1+x)^{11} \): The term with \( x^{11} \) is \( \beta \binom{11}{9}x^{11} \). The coefficient of \( x^{11} \) is: \[ C_{11} = \binom{11}{11} + \alpha \binom{11}{10} + \beta \binom{11}{9} \] \[ \binom{11}{11} = 1, \binom{11}{10} = 11, \binom{11}{9} = 55 \] Given \( C_{11} = 144 \): \[ 1 + 11\alpha + 55\beta = 144 \Rightarrow 11\alpha + 55\beta = 143 \] Divide by 11: \[ \alpha + 5\beta = 13 \text{(Equation 2)} \] Now we have the system of equations: \begin{align} \alpha + 3\beta &= 7 \text{(1)}
\alpha + 5\beta &= 13 \text{(2)} \end{align} Subtract Equation (1) from Equation (2): \[ (\alpha + 5\beta) - (\alpha + 3\beta) = 13 - 7 \Rightarrow 2\beta = 6 \Rightarrow \beta = 3 \] Substitute \( \beta = 3 \) into Equation (1): \[ \alpha + 3(3) = 7 \Rightarrow \alpha + 9 = 7 \Rightarrow \alpha = -2 \] Finally, we need to find \( \alpha^2 + \beta^2 \): \[ \alpha^2 + \beta^2 = (-2)^2 + (3)^2 = 4 + 9 = 13 \]