NTA Abhyas JEE Main2020MathematicsComplex NumberPractice
The value of ∑ k = 1 99 i k ! + ω k ! is (where, i = - 1 and ω is non-real cube root of unity)
Options
- A190 + ω
- B192 + ω 2
- C190 + i
- D192 + i
Correct answer
C. 190 + i
Step-by-step solution
∑ k = 1 99 i k ! + ∑ k = 1 99 ω k ! ∑ k = 1 99 i k ! = i 1 ! + i 2 ! + i 3 ! + i 4 ! + … + i 99 ! = i - 1 + i 6 + 1 + 1 + 1 + … + 1 = i - 2 + 96 = i + 94 ∑ k = 1 99 ω k ! = ω 1 + ω 2 ! + ω 3 ! + ω 4 ! + … + ω 99 ! = ω + ω 2 + 1 + 1 + 1 + … + 1 = 96 Sum = i + 94 + 96 = i + 190