the value of `((i^(11)+i^(12)+i^(13)+i^(14) +i^(15)))/((1+i))`is

Options

A. `(-(1+i))/2`
B. `((1-i))/2`
C. ` ((1+i))/2`
D. `1/2`

Answer

To solve the expression \(\frac{i^{11} + i^{12} + i^{13} + i^{14} + i^{15}}{1 + i}\), we will follow these steps:

### Step 1: Calculate the powers of \(i\)

The powers of \(i\) cycle every four terms:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

Using this cycle, we can calculate:
- \(i^{11} = i^{4 \cdot 2 + 3} = i^3 = -i\)
- \(i^{12} = i^{4 \cdot 3} = 1\)
- \(i^{13} = i^{4 \cdot 3 + 1} = i^1 = i\)
- \(i^{14} = i^{4 \cdot 3 + 2} = i^2 = -1\)
- \(i^{15} = i^{4 \cdot 3 + 3} = i^3 = -i\)

Now, we can sum these values:
\[ i^{11} + i^{12} + i^{13} + i^{14} + i^{15} = (-i) + 1 + i + (-1) + (-i) \]

### Step 2: Simplify the sum

Combining the terms:
\[ (-i + i - i) + (1 - 1) = -i \]

So, we have:
\[ i^{11} + i^{12} + i^{13} + i^{14} + i^{15} = -i \]

### Step 3: Substitute into the original expression

Now substituting back into the original expression:
\[ \frac{-i}{1 + i} \]

### Step 4: Rationalize the denominator

To simplify \(\frac{-i}{1 + i}\), we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{-i(1 - i)}{(1 + i)(1 - i)} = \frac{-i + i^2}{1^2 - i^2} = \frac{-i - 1}{1 - (-1)} = \frac{-i - 1}{2} \]

### Step 5: Final simplification

This can be rewritten as:
\[ \frac{-1 - i}{2} = -\frac{1}{2} - \frac{i}{2} \]

Thus, the final value of the expression \(\frac{i^{11} + i^{12} + i^{13} + i^{14} + i^{15}}{1 + i}\) is:
\[ -\frac{1}{2} - \frac{i}{2} \]