The value of $\displaystyle\sum_{n=1}^{13}\left(i^n+i^{n-1}\right)$ is : (a) $1$ (b) $1+i$ (c) $0$ (d) $i$

> [!TLDR] > Factoring out $i^{n-1}$ turns the sum into $(i+1)$ times a sum of consecutive powers of $i$, which collapses using the period-4 cycle of $i^m$. > > 1. $i^n+i^{n-1}=i^{n-1}\cdot i+i^{n-1}=i^{n-1}(i+1)$. > 2. So $\displaystyle\sum_{n=1}^{13}(i^n+i^{n-1})=(i+1)\sum_{n=1}^{13}i^{n-1}=(i+1)\sum_{m=0}^{12}i^m$ (substituting $m=n-1$). > 3. Powers of $i$ cycle with period $4$: $i^0+i^1+i^2+i^3=1+i-1-i=0$. …

The value of Σn=1¹³(iⁿ+iⁿ⁻¹) is : (a) 1 (b) 1+i (c) 0 (d) i

Question 115 of 122

\displaystyle\sum_{n=1}^{13}\left(i^n+i^{n-1}\right)

(a)

1

(b)

1+i

(c)

0

(d)

i

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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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Factoring out $i^{n-1}$ turns the sum into $(i+1)$ times a sum of consecutive powers of $i$ , which collapses using the period-4 cycle of $i^m$ .

$i^n+i^{n-1}=i^{n-1}\cdot i+i^{n-1}=i^{n-1}(i+1)$ .
So $\displaystyle\sum_{n=1}^{13}(i^n+i^{n-1})=(i+1)\sum_{n=1}^{13}i^{n-1}=(i+1)\sum_{m=0}^{12}i^m$ (substituting $m=n-1$ ).
Powers of $i$ cycle with period $4$ : $i^0+i^1+i^2+i^3=1+i-1-i=0$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 94 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\sqrt{5} + 3i$
(a) $\sqrt{5} - 3i$
(b) $\dfrac{\sqrt{5}}{14} + i \cdot \dfrac{3}{14}$
(c) $\dfrac{\sqrt{5}}{14} - i \cdot \dfrac{3}{14}$
(d) $3 - \sqrt{5}i$
›Reveal solutionSolution

The multiplicative inverse of $a+ib$ is $\dfrac{a-ib}{a^2+b^2}$ ; here $a=\sqrt5, b=3$ .

The multiplicative inverse of $z = \sqrt5 + 3i$ is $\dfrac{1}{z}$ . Rationalise by multiplying by the conjugate $\sqrt5 - 3i$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{1+i}{1-i} - \dfrac{1-i}{1+i}$
(a) 2
(b) -2
(c) 1
(d) -1
›Reveal solutionSolution

Simplify each fraction using the conjugate, subtract, then take the modulus of the resulting purely imaginary number.

$\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{(1-i)(1+i)} = \dfrac{1+2i+i^2}{1-i^2} = \dfrac{1+2i-1}{1+1} = \dfrac{2i}{2} = i$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\left(\dfrac{1+i}{1-i}\right)^m = 1$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

Simplify the base to $i$ , then find the smallest positive integer power of $i$ equal to 1 (the order of $i$ is 4).

As shown above, $\dfrac{1+i}{1-i} = i$ . So the equation becomes $i^m = 1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$i^{28}$
(a) $i$
(b) $-i$
(c) $1$
(d) $-1$
›Reveal solutionSolution

$i^{28} = 1$ , option (c).

Recall $i^2=-1$ , so $i^4 = (i^2)^2 = (-1)^2 = 1$ . Powers of $i$ cycle with period 4: $i, -1, -i, 1, i, -1, \dots$

…

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CBSE 2026Set ANNUAL1 mark
$2 - 3i$
›Reveal solutionSolution

The conjugate of $2-3i$ is $2+3i$ .

For a complex number $z=a+ib$ , the conjugate $\bar z = a - ib$ — the sign of the imaginary part is reversed while the real part stays unchanged.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Multiplicative inverse of 1 + \sqrt3i is:$
(a) 1 - √3i
(b) (1 - √3i)/2
(c) (1 + √3i)/4
(d) None of these
›Reveal solutionSolution

z⁻¹ = conjugate(z)/|z|²; for z = 1+√3i this gives (1−√3i)/4, which isn't among options (a)-(c) as written, so the correct choice is (d).

For a complex number $z = 1 + \sqrt{3}i$ , the multiplicative inverse is:

$z^{-1} = \frac{\bar z}{|z|^2}$

Here $\bar z = 1 - \sqrt{3}i$ , and $|z|^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4$ .

So:

$z^{-1} = \frac{1-\sqrt3 i}{4}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. The modulus of (1 + i)/(1 - i) is ..............$
›Reveal solutionSolution

Rationalize (1+i)/(1-i) to a+bi form, or use |z1/z2| = |z1|/|z2| — both give modulus 1.

Method using the modulus quotient rule $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ :

$|1+i| = \sqrt{1^2+1^2}=\sqrt2, \qquad |1-i|=\sqrt{1^2+(-1)^2}=\sqrt2$

$\left|\frac{1+i}{1-i}\right| = \frac{\sqrt2}{\sqrt2} = 1$

…

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CBSE 2026Set ANNUAL1 markMCQ
$i^9 \cdot i^{19}$
(a) $-1$
(b) $-i$
(c) $1$
(d) $0$
›Reveal solutionSolution

Using $i^4=1$ , reduce the exponents mod 4: $i^9 = i$ and $i^{19}=i^3=-i$ , so the product is $i\cdot(-i) = 1$ .

Recall $i^2=-1$ , and powers of $i$ repeat with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ .

$i^9 = i^{4\times2+1} = (i^4)^2 \cdot i = 1\cdot i = i$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$3 + 5i$
(a) $-3+5i$
(b) $-3-5i$
(c) $3-5i$
(d) $3+i$
›Reveal solutionSolution

The conjugate of $z=a+ib$ is $\bar{z}=a-ib$ ; only the sign of the imaginary part changes.

Here $z = 3+5i$ , so $a=3$ , $b=5$ .

…

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CBSE 2026Set ANNUAL1 mark
$i^9 + i^{19}$
›Reveal solutionSolution

Using $i^4=1$ , reduce each exponent mod 4: $i^9=i$ and $i^{19}=-i$ , which sum to 0.

Since $i^4=1$ , any power $i^k = i^{(k \bmod 4)}$ .

$i^9 = i^{(4\times2+1)} = (i^4)^2\cdot i^1 = 1\cdot i = i$ .

…

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CBSE 2026Set ANNUAL1 mark
$5 - 3i$
›Reveal solutionSolution

Modulus of $a+ib$ is $\sqrt{a^2+b^2}$ ; here that gives $\sqrt{34}$ .

For $z=5-3i$ , $a=5$ , $b=-3$ .

…

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CBSE 2026Set 1A1 markMCQ
$\sqrt{-25} \times \sqrt{-9}$
(1) $15$
(2) $-15$
(3) $15i$
(4) None of these
›Reveal solutionSolution

Write $\sqrt{-25}=5i$ and $\sqrt{-9}=3i$ ; their product is $15i^2=-15$ .

The rule $\sqrt{a}\,\sqrt{b}=\sqrt{ab}$ fails for negative numbers, so convert first:

$\sqrt{-25}=5i,\qquad \sqrt{-9}=3i.$ …

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Q.The value of ∑n=113(in+in−1)\displaystyle\sum_{n=1}^{13}\left(i^n+i^{n-1}\right)n=1∑13​(in+in−1) is :

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Q.The value of $\displaystyle\sum_{n=1}^{13}\left(i^n+i^{n-1}\right)$ is :

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