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Note 

i^2  = -1     

And -1 raised to an even power = 1

And -1 raised to an odd power  =  - 1

i^11 =  I^10 * i     =   (i^2)^5 * i      =  (-1)^5 * i   = (-1) * i  =  - i

i^16  = (i^2)^8   = (-1)^8  =  1

i^21  = i^20 * i   =  (i^2)^10 * i  =  (-1)^10 * i   = 1 * i   =   i

i^26  = (i^2)^13  =  (-1)^13  = - 1

i^31  =  i^30 * i  =  (i^2)^15 * i   = (-1)^15 * i  =  -1 * i  = - i

So....adding the 5  results we get

-i + 1+ i - 1 - i  =

 - i

EDIT MADE....THX Pushy!!

So 

i^1 = i,

i ^ 2 = -1,

i^3 = -i,

i^4= 1,

i ^ 5 = i,

and it keeps repeating between the four (i,-1,-i,i).

So, to find i^11, we find 11mod4 (or the remainder when dividing 11 and 4) to get 3. So, i ^11 is -i.

Then you do the rest.

i^16 = 1

i^21 = i

i^26 = -1

i^31 = -i.

So, all of those added together is -i + 1 + i - 1 - i which is -i.

Evaluate

\(i^{11} + i^{16} + i^{21} + i^{26} + i^{31}\)

Formula: \(i^{4n+a} = i^a\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ i^{11} + i^{16} + i^{21} + i^{26} + i^{31} } \\ &=& i^{4\cdot 2+3} + i^{4\cdot 4+0} + i^{4\cdot 5+1} + i^{4\cdot 6+2} + i^{4\cdot 7+3} \\ &=& i^{3} + i^{0} + i^{1} + i^{2} + i^{3} \\ &=& i^{2}i + 1 + i + i^{2} + i^{2}i \quad | \quad i^2 = -1 \\ &=& -i + 1 + i -1 + -i \\ &=& \mathbf{ -i } \\ \hline \end{array}\)