Question:

If $i = \sqrt{-1}$ and $n$ is a positive integer, then $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ is equal to

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The powers of $i$ repeat every four terms: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. This periodicity can help simplify complex expressions involving powers of $i$.

Updated On: May 3, 2025

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$i$
$i^n$
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The Correct Option is D

Approach Solution - 1

To solve the problem, we first consider the nature of $i = \sqrt{-1}$. We know that:

$i^1 = i$
$i^2 = -1$
$i^3 = -i$
$i^4 = 1$

These powers repeat every four terms, creating a cycle: $i, -1, -i, 1$. Now, let's analyze the expression $i^n + i^{n+1} + i^{n+2} + i^{n+3}$. Given the cyclic nature of powers of $i$:

$i^n$ could be any of $i, -1, -i, 1$.
$i^{n+1}$ follows $i^n$ in this order.
$i^{n+2}$ follows $i^{n+1}$, and so on.

Consequently, the expression $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ will cycle through all elements in one complete cycle:

If $i^n = i$, then the sum is $i - 1 - i + 1 = 0$.
If $i^n = -1$, the sum is $-1 - i + 1 + i = 0$.
If $i^n = -i$, the sum is $-i + 1 + i - 1 = 0$.
If $i^n = 1$, the sum is $1 + i - 1 - i = 0$.

In all cases, the sum $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$. Therefore, the correct answer is 0.

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Approach Solution -2

The problem involves working with powers of the imaginary unit $i = \sqrt{-1}$. The key property of powers of $i$ is their cyclical pattern:

$i^1 = i$
$i^2 = -1$
$i^3 = -i$
$i^4 = 1$

This pattern repeats every four exponents. Therefore, depending on $n \mod 4$, we can express any power of $i$ as:

$i^{4k} = 1$
$i^{4k+1} = i$
$i^{4k+2} = -1$
$i^{4k+3} = -i$

We are tasked with finding the sum $i^n + i^{n+1} + i^{n+2} + i^{n+3}$. Regardless of $n \mod 4$, observe that each sequence covers a complete cycle of four consecutive terms:

If $n \equiv 0 \mod 4$:
$(i^n, i^{n+1}, i^{n+2}, i^{n+3}) = (1, i, -1, -i)$
If $n \equiv 1 \mod 4$:
$(i^n, i^{n+1}, i^{n+2}, i^{n+3}) = (i, -1, -i, 1)$
If $n \equiv 2 \mod 4$:
$(i^n, i^{n+1}, i^{n+2}, i^{n+3}) = (-1, -i, 1, i)$
If $n \equiv 3 \mod 4$:
$(i^n, i^{n+1}, i^{n+2}, i^{n+3}) = (-i, 1, i, -1)$

In each case, adding the four numbers results in zero:

For $n \equiv 0 \mod 4$: $1 + i - 1 - i = 0$
For $n \equiv 1 \mod 4$: $i - 1 - i + 1 = 0$
For $n \equiv 2 \mod 4$: $-1 - i + 1 + i = 0$
For $n \equiv 3 \mod 4$: $-i + 1 + i - 1 = 0$

Thus, the expression $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ simplifies to $0$ for any positive integer $n$.