Question

Let $(1 + x + x^2)^{20} = \sum_{k=0}^{40} a_k x^k$. If $\alpha a_{11} = \beta a_{10} + \gamma a_9$, where $\alpha, \beta, \gamma \in N$, then the value of $\frac{\gamma - \alpha}{\beta}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

To find the relation between the coefficients $a_{11}, a_{10}, a_9$ of the expansion $(1 + x + x^2)^{20} = \sum_{k=0}^{40} a_k x^k$, we use a technique involving differentiation.

Let $P(x) = (1 + x + x^2)^{20}$.
We can write $P(x) = \sum_{k=0}^{40} a_k x^k$.

Differentiate $P(x)$ with respect to $x$:
$P'(x) = 20(1 + x + x^2)^{19}(1 + 2x)$.

Now, multiply $P'(x)$ by $(1 + x + x^2)$:
$P'(x)(1 + x + x^2) = 20(1 + x + x^2)^{19}(1 + 2x)(1 + x + x^2)$
$P'(x)(1 + x + x^2) = 20(1 + 2x)(1 + x + x^2)^{20}$
Substitute $P(x)$ back into the equation:
$P'(x)(1 + x + x^2) = 20(1 + 2x)P(x)$.

Now, substitute the series forms of $P(x)$ and $P'(x)$ into this equation:
Since $P(x) = \sum_{k=0}^{40} a_k x^k$, then $P'(x) = \sum_{k=1}^{40} k a_k x^{k-1}$.

The left-hand side (LHS) becomes:
$(\sum_{k=1}^{40} k a_k x^{k-1})(1 + x + x^2) = \sum_{k=1}^{40} k a_k x^{k-1} + \sum_{k=1}^{40} k a_k x^k + \sum_{k=1}^{40} k a_k x^{k+1}$.

The right-hand side (RHS) becomes:
$20(1 + 2x)(\sum_{k=0}^{40} a_k x^k) = 20 \sum_{k=0}^{40} a_k x^k + 40 \sum_{k=0}^{40} a_k x^{k+1}$.

Now, we equate the coefficients of $x^{10}$ on both sides.

Coefficient of $x^{10}$ on LHS:
From $\sum_{k=1}^{40} k a_k x^{k-1}$: set $k-1=10 \Rightarrow k=11$. The term is $11 a_{11} x^{10}$.
From $\sum_{k=1}^{40} k a_k x^k$: set $k=10$. The term is $10 a_{10} x^{10}$.
From $\sum_{k=1}^{40} k a_k x^{k+1}$: set $k+1=10 \Rightarrow k=9$. The term is $9 a_9 x^{10}$.
So, the coefficient of $x^{10}$ on LHS is $11 a_{11} + 10 a_{10} + 9 a_9$.

Coefficient of $x^{10}$ on RHS:
From $20 \sum_{k=0}^{40} a_k x^k$: set $k=10$. The term is $20 a_{10} x^{10}$.
From $40 \sum_{k=0}^{40} a_k x^{k+1}$: set $k+1=10 \Rightarrow k=9$. The term is $40 a_9 x^{10}$.
So, the coefficient of $x^{10}$ on RHS is $20 a_{10} + 40 a_9$.

Equating the coefficients of $x^{10}$:
$11 a_{11} + 10 a_{10} + 9 a_9 = 20 a_{10} + 40 a_9$.

Rearrange the terms to match the given form $\alpha a_{11} = \beta a_{10} + \gamma a_9$:
$11 a_{11} = (20 - 10)a_{10} + (40 - 9)a_9$
$11 a_{11} = 10 a_{10} + 31 a_9$.

Comparing this with $\alpha a_{11} = \beta a_{10} + \gamma a_9$, we identify the values:
$\alpha = 11$
$\beta = 10$
$\gamma = 31$

Given that $\alpha, \beta, \gamma \in N$, these values satisfy the condition.

Finally, we need to find the value of $\frac{\gamma - \alpha}{\beta}$:
$\frac{\gamma - \alpha}{\beta} = \frac{31 - 11}{10} = \frac{20}{10} = 2$.

The final answer is $\boxed{2}$.

Explanation of the solution:
The problem involves finding a recurrence relation between coefficients of a polynomial expansion.

1. Define $P(x) = (1 + x + x^2)^{20} = \sum_{k=0}^{40} a_k x^k$.
2. Differentiate $P(x)$ to get $P'(x) = 20(1 + x + x^2)^{19}(1 + 2x)$.
3. Establish a relationship between $P(x)$ and $P'(x)$ by multiplying $P'(x)$ by $(1+x+x^2)$:
   $P'(x)(1+x+x^2) = 20(1+2x)P(x)$.
4. Substitute the series forms of $P(x)$ and $P'(x)$ into this equation:
   $(\sum_{k=1}^{40} k a_k x^{k-1})(1+x+x^2) = 20(1+2x)(\sum_{k=0}^{40} a_k x^k)$.
5. Equate the coefficients of $x^{10}$ on both sides of the equation.
   LHS coefficient of $x^{10}$: $11 a_{11} + 10 a_{10} + 9 a_9$.
   RHS coefficient of $x^{10}$: $20 a_{10} + 40 a_9$.
6. Set LHS coefficient equal to RHS coefficient: $11 a_{11} + 10 a_{10} + 9 a_9 = 20 a_{10} + 40 a_9$.
7. Rearrange the equation to match the given form $\alpha a_{11} = \beta a_{10} + \gamma a_9$:
   $11 a_{11} = 10 a_{10} + 31 a_9$.
8. Identify $\alpha=11, \beta=10, \gamma=31$.
9. Calculate the required expression: $\frac{\gamma - \alpha}{\beta} = \frac{31 - 11}{10} = \frac{20}{10} = 2$.