Complex Numbers Question - None

Question

The sequence S = i + 2i² + 3i³ + 4i⁴ + ... up to 100 terms simplifies to, where i = $\sqrt{-1}$

Accepted Answer

Solution:

$$S = i + 2i^2 + 3i^3 + \ldots + 100i^{100}$$

Multiply by i:

$$Si = i^2 + 2i^3 + \ldots + 99i^{100} + 100i^{101}$$

Subtract:

$$S(1-i) = i + i^2 + i^3 + \ldots + i^{100} - 100i^{101}$$

Group by sets of 4 (since $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for any integer n):

$$S(1-i) = (i + i^2 + i^3 + i^4) + (i^5 + i^6 + i^7 + i^8) + \ldots + (i^{97} + i^{98} + i^{99} + i^{100}) - 100i$$

$$S(1-i) = 0 + 0 + \ldots + 0 - 100i = -100i$$

$$S = \frac{-100i}{1-i} = \frac{-100i(1+i)}{(1-i)(1+i)} = \frac{-100i - 100i^2}{1 - i^2} = \frac{-100i + 100}{2} = \frac{100(1-i)}{2} = 50(1-i)$$

∴ Answer is (a)

Question: