The value of the sum n=1∑13(in+in+1), where i=−1, equals

We have i2=−1Thus, n=1∑4(in+in+1)=(i1+i2)+ (i2+i3)+ (i3+i4)+ (i4+i5) =(i−1)+(−1−i)+(−i+1)+(1+i)=0\Rightarrow n=1∑12(in+in+1)=0Now, only remains is i13+i14=i−1

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The value of the sum _{n=1}^{13}(i^n+i^{n+1}), where i= {-1},

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The value of the sum $n = 1 \sum 13 (i^{n} + i^{n + 1})$ , where $i = - 1$ , equals

i-1

-i

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Elizabeth Discussed

The value of the sum

n = 1 \sum 13 (i^{n} + i^{n + 1})

, where

i = - 1

, equals

12 mins ago

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12 mins ago

Text solutionVerified

We have

i^{2} = - 1

Thus,

n = 1 \sum 4 (i^{n} + i^{n + 1}) = (i^{1} + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) = (i - 1) + (- 1 - i) + (- i + 1) + (1 + i) = 0

\Rightarrow

n = 1 \sum 12 (i^{n} + i^{n + 1}) = 0

Now, only remains is

i^{13} + i^{14} = i - 1

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Question Text	The value of the sum $n = 1 \sum 13 (i^{n} + i^{n + 1})$ , where $i = - 1$ , equals
Answer Type	Text solution:1
Upvotes	150

The value of the sum n=1∑13​(in+in+1), where i=−1​, equals

Text solutionVerified

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The value of the sum $n = 1 \sum 13 (i^{n} + i^{n + 1})$ , where $i = - 1$ , equals