Question

Let $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + ... + a_{20} x^{20} $. If $ (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k $, then k is equal to _______

Hint:

To find the sum of coefficients of odd or even powers in a polynomial expansion, substitute \( x = 1 \) and \( x = -1 \) into the polynomial and use the resulting equations. To find a specific coefficient, use the binomial theorem or multinomial theorem as appropriate.

Answer 1

Correct Answer: 239

Given \( (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \dots + a_{20} x^{20} \). 
Substitute \( x = 1 \): \[ (1 + 1 + 1)^{10} = a_0 + a_1 + a_2 + \dots + a_{20} \] \[ 3^{10} = a_0 + a_1 + a_2 + \dots + a_{20} \quad ...(i) \] Substitute \( x = -1 \): \[ (1 - 1 + (-1)^2)^{10} = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \] \[ 1^{10} = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \] \[ 1 = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \quad ...(ii) \] Subtracting (ii) from (i): \[ 3^{10} - 1 = 2(a_1 + a_3 + a_5 + \dots + a_{19}) \] \[ a_1 + a_3 + a_5 + \dots + a_{19} = \frac{3^{10} - 1}{2} = \frac{59049 - 1}{2} = \frac{59048}{2} = 29524 \] To find \( a_2 \), we consider the coefficient of \( x^2 \) in the expansion of \( (1 + x + x^2)^{10} \). 
Using the binomial expansion of \( (1 + (x + x^2))^{10} \): \[ (1 + (x + x^2))^{10} = \binom{10}{0} + \binom{10}{1}(x + x^2) + \binom{10}{2}(x + x^2)^2 + \dots \] \[ = 1 + 10(x + x^2) + 45(x^2 + 2x^3 + x^4) + \dots \] The coefficient of \( x^2 \) is \( a_2 = 10 \cdot 1 + 45 \cdot 1 = 10 + 45 = 55 \). Given \( (a_1 + a_3 + a_5 + \dots + a_{19}) - 11a_2 = 121k \). 
Substitute the values: \[ 29524 - 11(55) = 121k \] \[ 29524 - 605 = 121k \] \[ 28919 = 121k \] \[ k = \frac{28919}{121} = 239 \]