What is the value of the sumn=2∑11(in+in+1) where i=−1?

n=1∑11(in+in+1)i=−1Now i2=−1,i3=−i,i4=1,i5=i after this values repeatsi2+i3+i4+i5=0∴n=2∑11in=i11+i12=−i+1∴n=2∑11in+in+1=−2i.

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Question

What is the value of the sum
$n = 2 \sum 11 (i^{n} + i^{n + 1})$ where $i = - 1$ ?

i

2 i

- 2 i

1 + i

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Harper Discussed

What is the value of the sum

n = 2 \sum 11 (i^{n} + i^{n + 1})

where

i = - 1

8 mins ago

Discuss this question LIVE

8 mins ago

n = 1 \sum 11 (i^{n} + i^{n + 1}) i = - 1

Now

i^{2} = - 1, i^{3} = - i, i^{4} = 1, i^{5} = i

after this values repeats

i^{2} + i^{3} + i^{4} + i^{5} = 0

∴ n = 2 \sum 11 i^{n} = i^{11} + i^{12} = - i + 1

∴ n = 2 \sum 11 i^{n} + i^{n + 1} = - 2 i

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Question Text	What is the value of the sum $n = 2 \sum 11 (i^{n} + i^{n + 1})$ where $i = - 1$ ?
Answer Type	Text solution:1
Upvotes	150