Question
Asked Nov 12, 2019
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n          Sum= [i3 − (i − 1)3]
i=1
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Expert Answer

Step 1

Refer to the question we need to find the value of the provided expression as,

Sum=f-(i-1)']
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Sum=f-(i-1)']

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Step 2

Now we know that i= sqrt(-1), where I is the iota in the complex number.

i = 1
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i = 1

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Step 3

Now solve the expr...

Sum=i(i-1)]
(a-b)a -b-3ab+3b°a
(i-1)'=(P--3 +3)
As
i = 1
=1
P-(i-1)-(-1-3(-1)+3i)
--i-(-i-1+3+3i)
--i-(2+2i
=-2-3i
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Sum=i(i-1)] (a-b)a -b-3ab+3b°a (i-1)'=(P--3 +3) As i = 1 =1 P-(i-1)-(-1-3(-1)+3i) --i-(-i-1+3+3i) --i-(2+2i =-2-3i

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Calculus

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