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the value of the following ,where i= sqrt(-1).
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\[\sqrt{-1} = i\] it's and imaginary number
\[\sum_{n=1}^{13}(i^n+i^{n+1})=?\]
@Sjano my question is above stated
try spiting up the summation in to two summations over the addition, then use the geometric formula for the two sums.
which geometric form..
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if you have:\[\sum_{k=1}^nr^k=\frac{r^{n+1}-r}{r-1}\]i believe that should do the job.
should i do the summation fully upto 13 or i can get it by only common sense
(sqrt-1-1)+(-1-sqrt-1)+(-sqrt-1+1)+(1+sqrt-1)+(sqrt-1-1)+(-1-sqrt-1)+(-sqrt-1+1)+(1+sqrt-1)+(sqrt-1-1)+(-1-sqrt-1)+(-sqrt-1+1)+(1+sqrt-1)+(sqrt-1-1) =sqrt-1-1
I have done the same thing but isn't there a short trick. I mean i don't want to sum it fully.
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k thanx! i got it
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