Asked by Crow
                1+(2/i)+(3/i^2)+(4/i^3)+⋯+(n/i^(n−1))=29+28i
what is the value of n?
n is positive value
            
        what is the value of n?
n is positive value
Answers
                    Answered by
            Steve
            
    Well, since i^2 = -1 and i^3 = -i and i^4 = 1 and 1/i = -i, we have
1-2i-3+4i+5-6i-7+8i+9+...=29+28i
1-3+5-7+9-...-2i+4i-6i+8i-... = 29+28i
1+2(n/4) + 2i(n/4) = 29+28i
1+n/2 + n/2 i = 29+28i
Looks like n is about 56. That gives us
1+(-3+5)+...+(-55+57)
+(-2i+4i)+...+(-54+56) = -28+28i
Looks like we need to include 57 to get 29+28i
    
1-2i-3+4i+5-6i-7+8i+9+...=29+28i
1-3+5-7+9-...-2i+4i-6i+8i-... = 29+28i
1+2(n/4) + 2i(n/4) = 29+28i
1+n/2 + n/2 i = 29+28i
Looks like n is about 56. That gives us
1+(-3+5)+...+(-55+57)
+(-2i+4i)+...+(-54+56) = -28+28i
Looks like we need to include 57 to get 29+28i
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