Answer
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
Work Step by Step
(a)
4$\Sigma_{k=1}$ $(k-1)^{2}$=$(1-1)^{2}$+$(2-1)^{2}$+$(3-1)^{2}$+$(4-1)^{2}$=0+1+4+9-1
(b)
3$\Sigma_{k=1}$ $(k+1)^{2}$=$(-1+1)^{2}$+$(0+2)^{2}$+$(1+1)^{2}$+$(2+1)^{2}$+$(3+1)^{2}$=0+1+4+9+16
(c)
-1$\Sigma_{k=-3}$ $(k+1)^{2}$=$(-3)^{2}$+$(-2)^{2}$+$(-1)^{2}$=9+4+1
$(2+1)^{2}$(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
