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: \[ 29524 - 11(55) = 121k \] \[ 29524 - 605 = 121k \] \[ 28919 = 121k \] \[ k = \frac{28919}{121} = 239 \]

Answer 2

The problem provides a polynomial expansion \( (1 + x + x^2)^{10} = \sum_{r=0}^{20} a_r x^r \) and asks to find the value of \(k\) from the given relation \( (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k \).

Concept Used:

To solve this problem, we use two main concepts from the binomial theorem and polynomial expansions:

1. Sum of Coefficients: For a polynomial \( P(x) = \sum a_r x^r \), the sum of coefficients of odd powers of \(x\) (\(a_1 + a_3 + \ldots\)) can be found using the expression: \[ \sum a_{\text{odd}} = \frac{P(1) - P(-1)}{2} \]
2. Finding a Specific Coefficient: To find a specific coefficient like \(a_2\) (the coefficient of \(x^2\)), we can use the binomial expansion of the given expression, treating it as \((1 + (x+x^2))^{10}\), and collect the terms that produce \(x^2\). The general binomial expansion is \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Step-by-Step Solution:

Step 1: Find the sum of the odd coefficients \( S_{odd} = a_1 + a_3 + \ldots + a_{19} \).

Let \( P(x) = (1 + x + x^2)^{10} \). We evaluate \(P(x)\) at \(x=1\) and \(x=-1\).

\[ P(1) = (1 + 1 + 1^2)^{10} = 3^{10} \] \[ P(-1) = (1 - 1 + (-1)^2)^{10} = (1 - 1 + 1)^{10} = 1^{10} = 1 \]

The sum of odd coefficients is given by:

\[ S_{odd} = \frac{P(1) - P(-1)}{2} = \frac{3^{10} - 1}{2} \]

Step 2: Find the coefficient \(a_2\).

We need to find the coefficient of \(x^2\) in the expansion of \((1 + x + x^2)^{10}\). Let's expand it using the binomial theorem with \(a=1\) and \(b=x+x^2\):

\[ (1 + (x+x^2))^{10} = \binom{10}{0} + \binom{10}{1}(x+x^2)^1 + \binom{10}{2}(x+x^2)^2 + \binom{10}{3}(x+x^2)^3 + \ldots \]

We only need to consider the terms that can produce \(x^2\):

• From the term \(\binom{10}{1}(x+x^2)\): The coefficient of \(x^2\) is \(\binom{10}{1} = 10\).
• From the term \(\binom{10}{2}(x+x^2)^2 = \binom{10}{2}(x^2 + 2x^3 + x^4)\): The coefficient of \(x^2\) is \(\binom{10}{2} = \frac{10 \times 9}{2} = 45\).

Higher order terms like \(\binom{10}{k}(x+x^2)^k\) for \(k \ge 3\) will not produce an \(x^2\) term, as the lowest power of \(x\) is \(x^k\).

Therefore, the total coefficient of \(x^2\) is \(a_2\):

\[ a_2 = 10 + 45 = 55 \]

Step 3: Substitute the calculated values into the given equation.

The equation is \( (a_1 + a_3 + \ldots + a_{19}) - 11a_2 = 121k \).

Substituting the values of \(S_{odd}\) and \(a_2\):

\[ \frac{3^{10} - 1}{2} - 11(55) = 121k \]

Step 4: Solve the equation for \(k\).

First, calculate \(3^{10}\): \(3^{10} = (3^5)^2 = (243)^2 = 59049\).

Now, substitute this value back into the equation:

\[ \frac{59049 - 1}{2} - 605 = 121k \] \[ \frac{59048}{2} - 605 = 121k \] \[ 29524 - 605 = 121k \] \[ 28919 = 121k \]

Now, we solve for \(k\):

\[ k = \frac{28919}{121} \]

Since \(121 = 11^2\), we can perform the division:

\[ k = \frac{28919}{121} = 239 \]

The value of k is 239.