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Mathematics 12 Online
OpenStudy (maheshmeghwal9):

the value of the following ,where i= sqrt(-1).

OpenStudy (anonymous):

\[\sqrt{-1} = i\] it's and imaginary number

OpenStudy (maheshmeghwal9):

\[\sum_{n=1}^{13}(i^n+i^{n+1})=?\]

OpenStudy (maheshmeghwal9):

@Sjano my question is above stated

OpenStudy (anonymous):

try spiting up the summation in to two summations over the addition, then use the geometric formula for the two sums.

OpenStudy (maheshmeghwal9):

which geometric form..

OpenStudy (anonymous):

if you have:\[\sum_{k=1}^nr^k=\frac{r^{n+1}-r}{r-1}\]i believe that should do the job.

OpenStudy (maheshmeghwal9):

should i do the summation fully upto 13 or i can get it by only common sense

OpenStudy (ajprincess):

(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

OpenStudy (maheshmeghwal9):

I have done the same thing but isn't there a short trick. I mean i don't want to sum it fully.

OpenStudy (anonymous):

OpenStudy (maheshmeghwal9):

k thanx! i got it

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