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Evaluate ∑n=113(in+in+1), where n ∈ N.

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Question

Evaluate `sum_(n = 1)^13 (i^n + i^(n + 1))`, where n ∈ N.

Sum
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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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Chapter 5: Complex Numbers and Quadratic Equations - Exercise [Page 91]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 5 Complex Numbers and Quadratic Equations
Exercise | Q 2 | Page 91

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