Question

Compute i^1+i^2+i^3....i^99+i^100

Answer 1

Good morning ☕️

Answer:

i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰ = 0

Step-by-step explanation:

Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰

S =  i¹ +  i² +  i³ + . . . + i⁹⁹  +  i¹⁰⁰

S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀

then, S is the sum of 100 consecutive terms of a geometric sequence (an)

where the first term a1 = i¹ = i  and the common ratio = i

FORMULA:______________________

$S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}$

_______________________________

then

$S=i*\frac{1-i^{100} }{1-i}$

or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1  (we know that i⁴ = 1)

Hence

S = 0