The sum of the series $i+i^2+i^3+\ldots$ upto 1000 terms is _____.

> [!TLDR] > The powers of $i$ repeat in a cycle of four terms, and their sum over one cycle is zero. Since 1000 terms comprise exactly 250 such cycles, the total sum is $\boxed{0}$. Let's understand the fundamental property of the imaginary unit $i$ that makes this series easy to sum. The key lies in the cyclic nature of its powers. The imaginary unit $i$ is defined such that $i^2 = -1$. From this definition, all higher integer powers of $i$ follow a repeating pattern. This pattern is crucial for summing series involving powers of $i$. ### The Cyclic Nature of Powers of $i$ Consider the first few positive integer powers of $i$: * $i^1 = i$ * $i^2 = -1$ * $i^3 = i^2 \cdot i = (-1) \cdot i = -i$ * $i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1$ Now, let's look at the next set of powers: * $i^5 = i^4 \cdot i = 1 \cdot i = i$ * $i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$ * $i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i$ * $i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1$ Notice that the values $i, -1, -i, 1$ repeat every four powers. This is a cycle of length 4. > [!FORMULA] > For any integer $n$, the value of $i^n$ can be determined by the remainder when $n$ is divided by 4: > * If $n \equiv 1 \pmod{4}$, then $i^n = i$ > * If $n \equiv 2 \pmod{4}$, then $i^n = -1$ > * If $n \equiv 3 \pmod{4}$, then $i^n = -i$ > * If $n \equiv 0 \pmod{4}$, then $i^n = 1$ ### Sum of a Cycle An important consequence of this cycle is what happens when we sum four consecutive powers of $i$: $i^1 + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1$ $= (i - i) + (-1 + 1)$ $= 0 + 0 = 0$ > [!TIP] > The sum of any four consecutive integer powers of $i$ is always zero. For example, $i^{10} + i^{11} + i^{12} + i^{13} = (-1) + (-i) + 1 + i = 0$. Now, let's apply this understanding to the given series. ### Step-by-step Solution The series is $S = i+i^2+i^3+\ldots+i^{1000}$. 1. **Identify the number of terms:** The series has terms from $i^1$ to $i^{1000}$, so there are 1000 terms in total. 2

The sum of the series i+i²+i³+… upto 1000 terms is ___.

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i+i^2+i^3+\ldots

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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The powers of $i$ repeat in a cycle of four terms, and their sum over one cycle is zero. Since 1000 terms comprise exactly 250 such cycles, the total sum is $\boxed{0}$ .

Let's understand the fundamental property of the imaginary unit $i$ that makes this series easy to sum. The key lies in the cyclic nature of its powers.

The imaginary unit $i$ is defined such that $i^2 = -1$ . From this definition, all higher integer powers of $i$ follow a repeating pattern. This pattern is crucial for summing series involving powers of $i$ .

The Cyclic Nature of Powers of $i$

Consider the first few positive integer powers of $i$ :

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

Now, let's look at the next set of powers:

$i^5 = i^4 \cdot i = 1 \cdot i = i$
$i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$
$i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i$
$i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1$

Notice that the values $i, -1, -i, 1$ repeat every four powers. This is a cycle of length 4.

n

Sum of a Cycle

An important consequence of this cycle is what happens when we sum four consecutive powers of $i$ :

$i^1 + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1$

$= (i - i) + (-1 + 1)$

$= 0 + 0 = 0$

Tip

The sum of any four consecutive integer powers of $i$ is always zero. For example, $i^{10} + i^{11} + i^{12} + i^{13} = (-1) + (-i) + 1 + i = 0$ .

Now, let's apply this understanding to the given series.

Step-by-step Solution

The series is $S = i+i^2+i^3+\ldots+i^{1000}$ .

Identify the number of terms:

The series has terms from $i^1$ to $i^{1000}$ , so there are 1000 terms in total.
Group terms into cycles of four:

Since the sum of every four consecutive powers of $i$ is 0, we can group the terms of the series into sets of four.

$S = (i^1+i^2+i^3+i^4) + (i^5+i^6+i^7+i^8) + \ldots + (i^{997}+i^{998}+i^{999}+i^{1000})$
Determine the number of full cycles:

We have 1000 terms, and each cycle consists of 4 terms.

Number of cycles $= \frac{\text{Total number of terms}}{\text{Terms per cycle}} = \frac{1000}{4} = 250$ . …

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Q.The sum of the series i+i2+i3+…i+i^2+i^3+\ldotsi+i2+i3+… upto 1000 terms is _____.

The Cyclic Nature of Powers of iii

Sum of a Cycle

Step-by-step Solution

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Q.The sum of the series $i+i^2+i^3+\ldots$ upto 1000 terms is _____.

The Cyclic Nature of Powers of $i$

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