evaluate the sum

Question

(4 sigma k=0) ((-1)^k(k+3)²+4)

 

Answer 1

S = Σ{k=0,4}( (-1)^k (k+3)^2 + 4)

 

{k=0,4} means we want to use 0,1,2,3, and 4 for k.

 

(-1)^k alternates between +1 for even k and –1 for odd k.

 

k=0: ( (-1)^0 (0+3)^2 + 4) = (+9+4) = 13

 

k=1: ( (-1)^1 (1+3)^2 + 4) = (–16+4) = –12

 

k=2: ( (-1)^2 (2+3)^2 + 4) = (+25+4) = 29

 

k=3: ( (-1)^3 (3+3)^2 + 4) = (–36+4) = –32

 

k=4: ( (-1)^4 (4+3)^2 + 4) = (+49+4) = 53

 

S = 13 – 12 + 29 – 32 + 53

 

S = 1 + 29 + 21 = 30 + 21 = 51