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Evaluate the Sum of i^(k!) + omega^(k!) from k = 1 to 100 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

The value of

\sum_{k=1}^{100}\left(i^{\,k!} + \omega^{\,k!}\right),

where

i = \sqrt{-1}

and

\omega

is a complex cube root of unity, is

190 + \omega

192 + \omega^{2}

190 + i

192 + i

correct

Solution

Step 1: Split the sum into its two independent parts.

S = \sum_{k=1}^{100} i^{\,k!} + \sum_{k=1}^{100}\omega^{\,k!} .

Step 2: For the first sum, recall

i^{n}

depends only on

n \bmod 4

. So we need

k! \bmod 4

. Step 3: Once

k \ge 4

, the product

k!

contains both

2

and

4

, so

4 \mid k!

and

i^{\,k!} = i^{0} = 1

. Only

k = 1, 2, 3

need checking:

k=1:\ i^{1} = i, \qquad k=2:\ i^{2} = -1, \qquad k=3:\ i^{6} = i^{2} = -1 .

Step 4: There are

100 - 3 = 97

values of

k

from

4

100

, each contributing

1

\sum_{k=1}^{100} i^{\,k!} = i - 1 - 1 + 97 = i + 95 .

Step 5: For the second sum,

\omega^{n}

depends only on

n \bmod 3

. Once

k \ge 3

3 \mid k!

, so

\omega^{\,k!} = 1

. Only

k = 1, 2

differ:

k=1:\ \omega^{1} = \omega, \qquad k=2:\ \omega^{2} .

Step 6: There are

100 - 2 = 98

values of

k

from

3

100

, each giving

1

, and

\omega + \omega^{2} = -1

\sum_{k=1}^{100}\omega^{\,k!} = \omega + \omega^{2} + 98 = -1 + 98 = 97 .

Step 7: Add the two parts.

S = (i + 95) + 97 = 192 + i .

Answer: (4).

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