Q.47. Series summation

Question 47.
If i = √(-1), find Σ i^r  where r varies from r = (n + 6) to r = (9n + 84).

Answer 47.
Total number of terms
= (9n + 84) - (n + 5)
= 8n + 79
= 4(2n + 19) + 3
 As four consecutive terms add up to 0, we are left with 3 terms at the beginning
or at the end, sum of which will be the same.
=> S
= i^(n+6) + i^(n+7) + i^(n+8)
= i^n * (i^6 + i^7 + i^8)
= i^n * (-1 - i + 1)
= - i^(n+1) ... (1)
OR
= i^(n+1) / i^2
= i^(n+1-2)
= i^(n-1) ... (2)

From (2) and (1),
S = i^(n-1) OR - i^(n+1).

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