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Question
Evaluate `sum_(n = 1)^13 (i^n + i^(n + 1))`, where n ∈ N.
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Solution
We have `sum_(n = 1)^13 (i^n + i^(n + 1))`
= (i + i2) + (i2 + i3) + (i3 + i4) + (i4 + i5) + (i5 + i6) + (i6 + i7) + (i7 + i8) + (i8 + i9) + (i9 + i10) + (i10 + i11) + (i11 + i12) + (i12 + i13) + (i13 + i14)
= i + 2(i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 + i13) + i14
= i + 2[–1 – i + 1 + i – 1 – i + 1 + i – 1 – i + 1 + i] + (–1)
= i + 2(0) – 1
⇒ –1 + i
Hence, `sum_(n = 1)^13 (i^n + 1^(n + 1))` = –1 + i.
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