Sum of the series `sum_(r=0)^n (-1)^r ^nC_r[i^(5r)+i^(6r)+i^(7r)+i^(8r)]` is

Options

A. `2^(n)`
B. `2^(n//2+1`
C. `n^(n)+2^(n//2+1)`
D. `2^(n)+2^(n//2+1)cos(npi)/4`

Answer

To solve the given series

\[ S = \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( i^{5r} + i^{6r} + i^{7r} + i^{8r} \right), \]

we will break it down step by step.

### Step 1: Simplifying the terms inside the summation

The expression inside the summation can be simplified by recognizing the powers of \(i\):

\[ i^{5r} + i^{6r} + i^{7r} + i^{8r} = i^{5r} + i^{6r} + i^{7r} + i^{8r}. \]

We can express the powers of \(i\) as follows:
- \(i^{5r} = i^{4r} \cdot i^r = (1) \cdot i^r = i^r\) (since \(i^4 = 1\))
- \(i^{6r} = i^{4r} \cdot i^{2r} = (1) \cdot (-1)^r = (-1)^r\)
- \(i^{7r} = i^{4r} \cdot i^{3r} = (1) \cdot (-i)^r = (-i)^r\)
- \(i^{8r} = i^{4r} \cdot i^{4r} = (1) \cdot (1) = 1\)

Thus,

\[ i^{5r} + i^{6r} + i^{7r} + i^{8r} = i^r + (-1)^r + (-i)^r + 1. \]

### Step 2: Substitute back into the summation

Now we can substitute this back into the original summation:

\[ S = \sum_{r=0}^{n} (-1)^r \binom{n}{r} \left( i^r + (-1)^r + (-i)^r + 1 \right). \]

This can be split into four separate summations:

\[ S = \sum_{r=0}^{n} (-1)^r \binom{n}{r} i^r + \sum_{r=0}^{n} (-1)^r \binom{n}{r} (-1)^r + \sum_{r=0}^{n} (-1)^r \binom{n}{r} (-i)^r + \sum_{r=0}^{n} (-1)^r \binom{n}{r} 1. \]

### Step 3: Evaluating each summation

1. **First Summation**:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} i^r = (1 - i)^n \quad \text{(by the Binomial Theorem)} \]

2. **Second Summation**:

\[ \sum_{r=0}^{n} (-1)^{2r} \binom{n}{r} = \sum_{r=0}^{n} \binom{n}{r} = 2^n. \]

3. **Third Summation**:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} (-i)^r = (1 + i)^n. \]

4. **Fourth Summation**:

\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} = (1 - 1)^n = 0. \]

### Step 4: Combine the results

Now we can combine the results of the four summations:

\[ S = (1 - i)^n + 2^n + (1 + i)^n + 0. \]

### Step 5: Simplifying further

Using the properties of complex numbers, we can simplify \( (1 - i)^n + (1 + i)^n \):

\[ (1 - i)^n = \sqrt{2}^n \left( \cos\left(\frac{n\pi}{4}\right) - i \sin\left(\frac{n\pi}{4}\right) \right), \]
\[ (1 + i)^n = \sqrt{2}^n \left( \cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right) \right). \]

Adding these two gives:

\[ (1 - i)^n + (1 + i)^n = 2\sqrt{2}^n \cos\left(\frac{n\pi}{4}\right). \]

### Final Result

Thus, the final result for the sum \(S\) is:

\[ S = 2^{n/2 + 1} \cos\left(\frac{n\pi}{4}\right) + 2^n. \]