What is the value of the sum `sum_(n=2)^(11) (i^(n)+i^(n+1)), where i=sqrt(-1)?`

Options

A. i
B. `2i`
C. `-2i`
D. `1+i`

Explanation

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{n=2}^{11} (i^n + i^{n+1}) \]

where \( i = \sqrt{-1} \).

### Step 1: Rewrite the sum
We can rewrite the sum as follows:

\[ S = \sum_{n=2}^{11} i^n + \sum_{n=2}^{11} i^{n+1} \]

The second sum can be adjusted to start from \( n=3 \) to \( n=12 \):

\[ S = \sum_{n=2}^{11} i^n + \sum_{n=3}^{12} i^n \]

### Step 2: Combine the sums
Now, we can combine the two sums:

\[ S = i^2 + (i^3 + i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^{10} + i^{11}) + i^{12} \]

### Step 3: Identify the powers of \( i \)
The powers of \( i \) cycle every 4 terms:

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \)
- \( i^6 = -1 \)
- \( i^7 = -i \)
- \( i^8 = 1 \)
- \( i^9 = i \)
- \( i^{10} = -1 \)
- \( i^{11} = -i \)
- \( i^{12} = 1 \)

### Step 4: Substitute the values
Now we can substitute these values into our expression for \( S \):

\[ S = (-1) + (2 \cdot (-i) + 2 \cdot 1 + 2 \cdot i + 2 \cdot (-1) + 2 \cdot (-i) + 1) \]

### Step 5: Simplify the expression
Calculating the contributions:

- From \( i^2 \): \( -1 \)
- From \( i^3 \): \( 2 \cdot (-i) = -2i \)
- From \( i^4 \): \( 2 \cdot 1 = 2 \)
- From \( i^5 \): \( 2 \cdot i = 2i \)
- From \( i^6 \): \( 2 \cdot (-1) = -2 \)
- From \( i^7 \): \( 2 \cdot (-i) = -2i \)
- From \( i^{12} \): \( 1 \)

Combining these:

\[ S = -1 + (-2i) + 2 + (2i) + (-2) + (-2i) + 1 \]

### Step 6: Combine like terms
Now, combine the real and imaginary parts:

Real part:
\[ -1 + 2 - 2 + 1 = 0 \]

Imaginary part:
\[ -2i + 2i - 2i = -2i \]

### Final Result
Thus, the value of the sum is:

\[ S = -2i \]