Maths · Algebra of complex numbers
The sequence upto 100 times simplifies to where
The sequence \( \boldsymbol{S}=\boldsymbol{i}+\boldsymbol{2} \boldsymbol{i}^{2}+\boldsymbol{3} \boldsymbol{i}^{3}+\ldots \ldots \) upto 100 times simplifies to where \( i= \) \( \sqrt{-1} \)
- A. \( 50(1-i) \)
- B. 25
- C. \( 25(1+i) \)
- D. \( 100(1-i) \)
Step-by-step solution
The sequence consists of terms n * i^n for n = 1 to 100. Since i^4 = 1, the powers repeat every 4. Summing each block of 4 consecutive terms: (1*i + 2*(-1) + 3*(-i) + 4*1) = 2 - 2i = 2(1-i). There are 25 such blocks (since 100/4=25), so total sum = 25 * 2(1-i) = 50(1-i).
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