Summand Simplification using Powers of i

The problem asks us to find the value of the following sum:

\( \sum_{n=1}^{20} (i^{n-1} + i^n + i^{n+1}) \) Here, \(i\) represents the imaginary unit, which is defined as \(i = \sqrt{-1}\).

To solve this, we can first simplify the general term within the summation. Let's call this term \(a_n\):

\( a_n = i^{n-1} + i^n + i^{n+1} \)

We can factor out the term with the lowest exponent, which is \(i^{n-1}\):

\( a_n = i^{n-1} (1 + i^1 + i^2) \)

Now, let's recall the fundamental properties of the powers of the imaginary unit \(i\):

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

These powers follow a cycle of 4. Using these properties, we can substitute the values into our expression for \(a_n\):

\( a_n = i^{n-1} (1 + i + (-1)) \)

Simplifying the expression inside the parentheses:

\( a_n = i^{n-1} (i) \)

Using the laws of exponents, specifically \(x^a \times x^b = x^{a+b}\), we get:

\( a_n = i^{(n-1) + 1} \) \( a_n = i^n \)

So, the original sum simplifies to summing just the powers of \(i\) from \(n=1\) to \(n=20\):

\( \sum_{n=1}^{20} i^n \)

Series Evaluation with Powers of i Cycle

Our task now is to evaluate the sum \(S = \sum_{n=1}^{20} i^n\). This means we need to calculate:

\( S = i^1 + i^2 + i^3 + i^4 + i^5 + \dots + i^{19} + i^{20} \)

Let's examine the sum of the first four consecutive powers of \(i\):

\( i^1 + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1 = 0 \)

This demonstrates a key property: the sum of any four consecutive powers of \(i\) is always zero.

The summation runs from \(n=1\) to \(n=20\). Since \(20\) is a multiple of \(4\) (\(20 = 4 \times 5\)), we can group the terms in the sum into exactly 5 sets, where each set contains four consecutive powers of \(i\).

\( S = (i^1 + i^2 + i^3 + i^4) + (i^5 + i^6 + i^7 + i^8) + \dots + (i^{17} + i^{18} + i^{19} + i^{20}) \)

Each of these groups sums to 0. For example, the second group:

\( i^5 + i^6 + i^7 + i^8 = i^4(i^1 + i^2 + i^3 + i^4) = 1 \times (0) = 0 \)

Since there are 5 such groups, and each sums to 0, the total sum \(S\) is:

\( S = 0 + 0 + 0 + 0 + 0 \) \( S = 0 \)

Final Answer Determination

After simplifying the expression \((i^{n-1} + i^n + i^{n+1})\) to \(i^n\), we evaluated the sum \(\sum_{n=1}^{20} i^n\). We utilized the property that the sum of any four consecutive powers of the imaginary unit \(i\) is zero. Because the total number of terms in the sum (20) is a multiple of 4, the entire sum consists of groups that add up to zero.

Therefore, the value of the sum \(\sum_{n=1}^{20}(i^{n-1} + i^n + i^{n+1})\) is 0.