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KCET 2026 Mathematics Solved Paper

Question : 3 of 60

Marks: +1, -0

\sum_{n=1}^{4}(-1)^{2 n} \cdot i^{2 n}=

Solution:

Concept:

Simplify the general term using properties of powers of

-1

and

i

Explanation:

For any positive integer

n

, the power

2n

is always even.

Therefore,

(-1)^{2n} = 1

So the general term becomes:

(-1)^{2n} \cdot i^{2n} = 1 \cdot i^{2n} = i^{2n}

Now use

i^2 = -1

i^{2n} = (i^2)^n = (-1)^n

Thus the required sum is:

\sum_{n=1}^{4} (-1)^n

Substitute

n = 1, 2, 3, 4

(-1)^1 + (-1)^2 + (-1)^3 + (-1)^4

= -1 + 1 - 1 + 1 = 0

Answer:

The value of the sum is

0

Hence, the correct option is C.

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