If x > 0, then the number of positive real terms in the expansion of `(1 + ix)^(4n)` is

Options

A. n+1
B. n-1
C. n=2r+1
D. 2n

Answer

To find the number of positive real terms in the expansion of \((1 + ix)^{4n}\), we will follow these steps:

### Step 1: Understand the Expansion
The binomial expansion of \((1 + ix)^{4n}\) can be expressed as:
\[ \sum_{k=0}^{4n} \binom{4n}{k} (ix)^k \]
This expands to:
\[ \sum_{k=0}^{4n} \binom{4n}{k} i^k x^k \]

### Step 2: Identify Real Terms
In the expansion, the terms will be real when \(k\) is even. This is because:
- \(i^k\) is real when \(k\) is even (specifically, \(i^0 = 1\), \(i^2 = -1\), \(i^4 = 1\), etc.).
- \(i^k\) is imaginary when \(k\) is odd.

### Step 3: Determine the Values of \(k\)
The even values of \(k\) from \(0\) to \(4n\) are:
\[ k = 0, 2, 4, \ldots, 4n \]
This forms an arithmetic sequence where the first term \(a = 0\) and the common difference \(d = 2\).

### Step 4: Count the Number of Even Terms
To find the number of terms in this sequence, we can use the formula for the \(n\)-th term of an arithmetic sequence:
\[ k_n = a + (n-1)d \]
Setting \(k_n = 4n\):
\[ 4n = 0 + (n-1) \cdot 2 \]
Solving for \(n\):
\[ 4n = 2n - 2 \implies 2n = 2 \implies n = 1 \]
Thus, the last term corresponds to \(n = 2n + 1\).

### Step 5: Find Total Number of Terms
The number of even terms from \(0\) to \(4n\) can be calculated as follows:
- The sequence of even integers from \(0\) to \(4n\) can be expressed as:
\[ 0, 2, 4, \ldots, 4n \]
The number of terms is given by:
\[ \text{Number of terms} = \frac{4n - 0}{2} + 1 = 2n + 1 \]

### Conclusion
The number of positive real terms in the expansion of \((1 + ix)^{4n}\) is \(2n + 1\).