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`

Answer

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: Simplifying the Expression
We can rewrite the expression inside the summation:

\[ i^n + i^{n+1} = i^n (1 + i) \]

Thus, we can express the sum as:

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

### Step 2: Factoring Out the Constant
Since \(1 + i\) is a constant with respect to the summation, we can factor it out:

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

### Step 3: Evaluating the Inner Sum
Now we need to evaluate the sum \(\sum_{n=2}^{11} i^n\). The powers of \(i\) cycle every 4 terms:

- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)
- \(i^5 = i\) (and so on)

Thus, we can compute the terms from \(i^2\) to \(i^{11}\):

\[ \begin{align*} i^2 & = -1 \\ i^3 & = -i \\ i^4 & = 1 \\ i^5 & = i \\ i^6 & = -1 \\ i^7 & = -i \\ i^8 & = 1 \\ i^9 & = i \\ i^{10} & = -1 \\ i^{11} & = -i \\ \end{align*} \]

### Step 4: Summing the Terms
Now we sum these values:

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

Grouping the terms:

- The real parts: \((-1 + 1 - 1 + 1 - 1) = -1\)
- The imaginary parts: \((-i - i - i) = -3i\)

Thus, we have:

\[ \sum_{n=2}^{11} i^n = -1 - 3i \]

### Step 5: Final Calculation
Now substituting back into our expression for \(S\):

\[ S = (1 + i)(-1 - 3i) \]

Using the distributive property:

\[ S = -1 - 3i + (-1)i - 3i^2 \]

Since \(i^2 = -1\):

\[ S = -1 - 3i - i + 3 = 2 - 4i \]

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

\[ \boxed{2 - 4i} \]