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

Options

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

Answer

To find the value of \( a_{2n-1} \) in the expression \( (1 + 4x + 4x^2)^n = \sum_{r=0}^{2n} a_r x^r \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression \( 1 + 4x + 4x^2 \):
\[ 1 + 4x + 4x^2 = 1 + (2x)^2 + 2(2x) \]
This can be recognized as a perfect square:
\[ = (1 + 2x)^2 \]

### Step 2: Raise to the power of \( n \)
Now we raise this expression to the power of \( n \):
\[ (1 + 4x + 4x^2)^n = ((1 + 2x)^2)^n = (1 + 2x)^{2n} \]

### Step 3: Apply the Binomial Theorem
Using the Binomial Theorem, we expand \( (1 + 2x)^{2n} \):
\[ (1 + 2x)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} (2x)^r = \sum_{r=0}^{2n} \binom{2n}{r} 2^r x^r \]

### Step 4: Identify the coefficient \( a_{2n-1} \)
We need to find the coefficient of \( x^{2n-1} \), which corresponds to \( a_{2n-1} \):
\[ a_{2n-1} = \binom{2n}{2n-1} 2^{2n-1} \]

### Step 5: Simplify the coefficient
Now we simplify the expression for \( a_{2n-1} \):
\[ \binom{2n}{2n-1} = \frac{2n!}{(2n-1)! \cdot 1!} = 2n \]
Thus,
\[ a_{2n-1} = 2n \cdot 2^{2n-1} \]

### Step 6: Final expression
Putting it all together, we have:
\[ a_{2n-1} = 2n \cdot 2^{2n-1} = n \cdot 2^{2n} \]

### Conclusion
Therefore, the value of \( a_{2n-1} \) is:
\[ \boxed{n \cdot 2^{2n-1}} \]