## University Calculus: Early Transcendentals (3rd Edition)

a) $$\sum_{k=1}^{4}(k-1)^{2}=(1-1)^{2}+(2-1)^{2}+(3-1)^{2}+(4-1)^{2}=0^{2}+1^{2}+2^{2}+3^{2}=1+4+9$$ b) $$\sum_{k=-1}^{3}(k+1)^{2}=(-1+1)^{2}+(0+1)^{2}+(1+1)^{2}+(2+1)^{2}+(3+1)^{2}=0^{2}+1^{2}+2^{2}+3^{2}+4^{2}=1+4+9+16$$ c)$$\sum_{k=-3}^{-1}k^{2}=(-3)^{2}+(-2)^{2}+(-1)^{2}=9+4+1=1+4+9$$ (b) is not equivalent to the other two.