Understanding the Sum of Powers of i

The question asks us to find the value of the sum \( \sum\limits_{n=1}^{8n+7} i^n \), where \( i = \sqrt{-1} \). This is a sum of consecutive powers of the imaginary unit \( i \). Let's denote the upper limit of the summation as \( N \). So, we need to evaluate \( \sum\limits_{k=1}^{N} i^k \), where \( N = 8n+7 \).

Properties of Powers of i

The powers of \( i \) follow a repeating pattern:

• \( i^1 = i \)
• \( i^2 = -1 \)
• \( i^3 = i^2 \cdot i = -1 \cdot i = -i \)
• \( i^4 = (i^2)^2 = (-1)^2 = 1 \)
• \( i^5 = i^4 \cdot i = 1 \cdot i = i \)

The pattern \( i, -1, -i, 1 \) repeats every 4 powers.

Sum of Consecutive Powers of i

Let's look at the sum of consecutive powers of \( i \):

• Sum of 1 term: \( i^1 = i \)
• Sum of 2 terms: \( i^1 + i^2 = i - 1 \)
• Sum of 3 terms: \( i^1 + i^2 + i^3 = i - 1 - i = -1 \)
• Sum of 4 terms: \( i^1 + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0 \)

Since the sum of a cycle of 4 consecutive powers of \( i \) is 0, the value of the sum \( \sum\limits_{k=1}^{N} i^k \) depends on the remainder when \( N \) is divided by 4.

If \( N = 4q \), the sum is 0.

If \( N = 4q+1 \), the sum is \( i \).

If \( N = 4q+2 \), the sum is \( i-1 \).

If \( N = 4q+3 \), the sum is \( -1 \).

Analyzing the Upper Limit \( 8n+7 \)

The upper limit of the sum is \( N = 8n+7 \). We need to determine the form of \( N \) when divided by 4. We can rewrite \( 8n+7 \) as follows:

\( N = 8n+7 = 8n + 4 + 3 \)

We can factor out 4 from the terms \( 8n+4 \):

\( N = 4(2n) + 4(1) + 3 \)

\( N = 4(2n+1) + 3 \)

This shows that \( N = 8n+7 \) is always of the form \( 4q+3 \), where \( q = 2n+1 \). This is true for any integer value of \( n \) for which the sum is defined (typically \( n \ge 1 \) in this context for the upper limit to be positive).

Calculating the Sum \( \sum\limits_{n=1}^{8n+7} i^n \)

Since the upper limit \( N = 8n+7 \) is always of the form \( 4q+3 \), the sum \( \sum\limits_{k=1}^{N} i^k \) is equal to the sum of the first three terms in the cycle \( i^1 + i^2 + i^3 \), which is \( i - 1 - i = -1 \).

Therefore, \( \sum\limits_{n=1}^{8n+7} i^n = -1 \).

Conclusion

The value of the sum \( \sum\limits_{n=1}^{8n+7} i^n \) is determined by the remainder when the upper limit \( 8n+7 \) is divided by 4. Since \( 8n+7 = 4(2n+1) + 3 \), the remainder is always 3. A sum of powers of \( i \) from \( i^1 \) up to \( i^N \) where \( N \) has a remainder of 3 when divided by 4, is always \( -1 \).

Revision Table: Complex Numbers and i

Additional Information: General Sum of Powers of i

The sum of the first \( N \) powers of \( i \) starting from \( i^1 \) can be generalized. The sum \( S_N = \sum\limits_{k=1}^{N} i^k \) is a geometric series with first term \( a = i \) and common ratio \( r = i \). The sum of a geometric series is given by \( S_N = a \frac{r^N - 1}{r - 1} \) for \( r \neq 1 \).

Here, \( a=i \) and \( r=i \). So, \( S_N = i \frac{i^N - 1}{i - 1} \).

We can simplify the denominator by multiplying the numerator and denominator by the conjugate of \( i-1 \), which is \( -i-1 \):

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

So, \( S_N = i \frac{i^N - 1}{i - 1} = i (i^N - 1) \left( \frac{-i-1}{2} \right) = \frac{i(i^N - 1)(-i-1)}{2} \)

\( S_N = \frac{i(-i^{N+1} - i^N + i + 1)}{2} = \frac{-i^2 i^N - i^{N+1} + i^2 + i}{2} = \frac{-(-1) i^N - i^{N+1} - 1 + i}{2} \)

\( S_N = \frac{i^N - i^{N+1} - 1 + i}{2} \)

Let's check this formula for different values of \( N \pmod 4 \).

• If \( N=4q \), \( i^N = 1 \), \( i^{N+1} = i \). \( S_{4q} = \frac{1 - i - 1 + i}{2} = \frac{0}{2} = 0 \).
• If \( N=4q+1 \), \( i^N = i \), \( i^{N+1} = -1 \). \( S_{4q+1} = \frac{i - (-1) - 1 + i}{2} = \frac{i + 1 - 1 + i}{2} = \frac{2i}{2} = i \).
• If \( N=4q+2 \), \( i^N = -1 \), \( i^{N+1} = -i \). \( S_{4q+2} = \frac{-1 - (-i) - 1 + i}{2} = \frac{-1 + i - 1 + i}{2} = \frac{-2 + 2i}{2} = -1 + i \).
• If \( N=4q+3 \), \( i^N = -i \), \( i^{N+1} = 1 \). \( S_{4q+3} = \frac{-i - 1 - 1 + i}{2} = \frac{-2}{2} = -1 \).

The formula confirms the pattern based on \( N \pmod 4 \). In our problem, \( N = 8n+7 = 4(2n+1)+3 \), so \( N \) is of the form \( 4q+3 \). Thus, the sum is \( -1 \).