What is the coefficient of `x^(n)` in `(x^(2) + 2x)^(n-1)` ?

Options

A. `(n-1)2^(n-1)`
B. `(n-1)xx2^(n-1)`
C. `(n-1)2^(n)`
D. `n.2(n-1)`

Answer

To find the coefficient of \( x^n \) in the expression \( (x^2 + 2x)^{n-1} \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Identify the terms in the binomial expansion
The expression can be rewritten as:
\[ (x^2 + 2x)^{n-1} \]
According to the Binomial Theorem, the expansion of \( (a + b)^m \) is given by:
\[ \sum_{r=0}^{m} \binom{m}{r} a^{m-r} b^r \]
In our case, \( a = x^2 \), \( b = 2x \), and \( m = n-1 \).

### Step 2: Write the general term of the expansion
The general term \( T_r \) in the expansion is:
\[ T_r = \binom{n-1}{r} (x^2)^{n-1-r} (2x)^r \]
This simplifies to:
\[ T_r = \binom{n-1}{r} x^{2(n-1-r)} (2^r x^r) = \binom{n-1}{r} 2^r x^{2(n-1-r) + r} \]
This can be further simplified to:
\[ T_r = \binom{n-1}{r} 2^r x^{2n - 2 - r} \]

### Step 3: Set the exponent of \( x \) equal to \( n \)
We need to find the coefficient of \( x^n \). Therefore, we set the exponent equal to \( n \):
\[ 2n - 2 - r = n \]
Solving for \( r \):
\[ 2n - r - 2 = n \implies n - r = 2 \implies r = n - 2 \]

### Step 4: Substitute \( r \) back into the general term
Now we substitute \( r = n - 2 \) back into the general term:
\[ T_{n-2} = \binom{n-1}{n-2} 2^{n-2} x^{n} \]
Using the property of combinations:
\[ \binom{n-1}{n-2} = n-1 \]
Thus, we have:
\[ T_{n-2} = (n-1) 2^{n-2} x^n \]

### Step 5: Identify the coefficient
The coefficient of \( x^n \) in the expansion is:
\[ (n-1) 2^{n-2} \]

### Final Answer
The coefficient of \( x^n \) in \( (x^2 + 2x)^{n-1} \) is:
\[ \boxed{(n-1) 2^{n-2}} \]