`"If" n in "and if "(1+ 4x +4 x^2)^n=underset(r=0)overset(2n)Sigma a_rx^r, "where" a_0,a_1,a_2,.....a_(2n) "are real number" `
The value of ` 2 ` underset(r=0) overset (n) Sigma_(2r) ` is

Options

A. `9^n-1`
B. `9 ^n+1`
C. `9^n-2`
D. `9^n +2`

Answer

To solve the problem, we need to analyze the expression given and derive the required summation. Here’s a step-by-step breakdown of the solution:

### Step 1: Rewrite the Expression
We start with the expression given:
\[ (1 + 4x + 4x^2)^n = \sum_{r=0}^{2n} a_r x^r \]
We can rewrite the left-hand side:
\[ 1 + 4x + 4x^2 = (1 + 2x)^2 \]
Thus, we have:
\[ (1 + 2x)^2^n = (1 + 2x)^{2n} \]

### Step 2: Expand Using Binomial Theorem
Using the Binomial Theorem, we can expand \((1 + 2x)^{2n}\):
\[ (1 + 2x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (2x)^k = \sum_{k=0}^{2n} \binom{2n}{k} 2^k x^k \]
This means:
\[ a_k = \binom{2n}{k} 2^k \]

### Step 3: Calculate the Required Summation
We need to find:
\[ 2 \sum_{r=1}^{n} a_{2r-1} \]
This means we need to evaluate the coefficients \(a_{2r-1}\) for \(r = 1\) to \(n\).

### Step 4: Identify Odd Coefficients
The odd coefficients in the expansion correspond to \(k = 1, 3, 5, \ldots, 2n-1\). Therefore, we can express the summation as:
\[ \sum_{r=1}^{n} a_{2r-1} = \sum_{k=1, k \text{ odd}}^{2n} \binom{2n}{k} 2^k \]

### Step 5: Use the Binomial Identity
We can use the identity for the sum of odd indexed binomial coefficients:
\[ \sum_{k \text{ odd}}^{m} \binom{m}{k} = 2^{m-1} \]
Thus, we have:
\[ \sum_{k \text{ odd}}^{2n} \binom{2n}{k} = 2^{2n-1} \]
So:
\[ \sum_{k \text{ odd}}^{2n} \binom{2n}{k} 2^k = 2 \cdot 2^{2n-1} = 2^{2n} \]

### Step 6: Final Calculation
Now, substituting back into our expression:
\[ 2 \sum_{r=1}^{n} a_{2r-1} = 2 \cdot 2^{2n} = 2^{2n + 1} \]

### Step 7: Relate to the Original Expression
From the original equation, we also know:
\[ (1 + 2x)^{2n} \text{ evaluated at } x = 1 \text{ gives } 3^{2n} = 9^n \]
Thus, we conclude:
\[ 2 \sum_{r=1}^{n} a_{2r-1} = 9^n - 1 \]

### Final Answer
The value of \(2 \sum_{r=1}^{n} a_{2r-1}\) is:
\[ 9^n - 1 \]