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  • Evaluate ∑_(n=1)^13▒(i^n+i^(n+1) ) where n ϵ N.

Evaluate \sum_{i=1}^{13}(i^n+i^{n+1} ) where n ϵ N.

Answers (1)

= \sum_{i=1}^{13}(i^n+i^{n+1} )

= \sum_{i=1}^{13}(1+i) i^n

=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )

=(1+i) \frac{i(i^{13}-1)}{(i-1)}

=(1+i) i\frac{(i-1)}{(i-1)}

=(1+i)i

=i+i^2

=i-1

 

Posted by

infoexpert21

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