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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 44 of 120

Marks: +1, -0

What is

i^{1000} + i^{1001} + i^{1002} + i^{1003}

equal to (where i

\sqrt{-1}

)?

0
i
-i
1

Solution:

Concept:

The powers of

i

repeat every 4 steps because

i^4=1

.

Explanation:

We need to find

i^{1000} + i^{1001} + i^{1002} + i^{1003}

.

Since

i^4 = 1

, each exponent can be reduced modulo

4

.

1000

is divisible by

4

(because

1000 = 4 \times 250

), so

i^{1000} = (i^4)^{250} = 1

.

1001 = 4 \times 250 + 1

, therefore

i^{1001} = i^1 = i

.

1002 = 4 \times 250 + 2

, therefore

i^{1002} = i^2 = -1

.

1003 = 4 \times 250 + 3

, therefore

i^{1003} = i^3 = -i

.

Now add the terms:

1 + i + (-1) + (-i) = 0

.

Answer:

0

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