Question

Find the sum of i^(1)+i^(2)+i^(3)+......+i^(99).

Answer 1

Answer:

• i¹ = i
• i² = -1
• i³ = -i
• i⁴ = 1

Now we know that,

⇒ i + i² + i³ + i⁴ = i - 1 - i + 1 = 0

Similarly this pattern continues for every 4 consecutive terms after this series.

⇒ i + i² + i³ + .... i⁹⁶ = 0

⇒ i⁹⁷ + i⁹⁸ + i⁹⁹ = i - 1 - i

⇒ i⁹⁷ + i⁹⁸ + i⁹⁹ = -1

Therefore the Sum of all terms from i to i⁹⁹ is -1.

Answer 2

$\huge\bf{Answer:-}$

[i¹ = i],

[i² = -1]

[i³ = -1]

[i^4 = 1]

Hence, [i + i² + i³ + i^4 = i + 1 + i + 1 = 0]

.•. 4 consecitive term

So,

[i + i² + i³ + till (i^96) = 0]

[i^97 + i^98 + i^99 = i - 1 - 1]

[i^97 + i^98 + i^99 = -1]

.•. {i^99 = -1}

__[Answer]