Answer
(a) $$\sum_{n=1}^5(1-n^2)=(1-1^2)+(1-2^2)+(1-3^2)+(1-4^2)+(1-5^2)=-50$$
(b) $$\sum_{n=3}^6(-1)^n2^{n-2}=(-1)^32^{1}+(-1)^42^{2}+(-1)^52^{3}+(-1)^62^{4}
=10$$
Work Step by Step
(a) The sigma notation means the sum of each term with $n=1,2,3,4,5$. Thus we have:
$$\sum_{n=1}^5(1-n^2)=(1-1^2)+(1-2^2)+(1-3^2)+(1-4^2)+(1-5^2)=0-3-8-15-24=-50$$
(b) Similarly for this case with $n=3,4,5,6$, we have:
$$\sum_{n=3}^6(-1)^n2^{n-2}=(-1)^32^{1}+(-1)^42^{2}+(-1)^52^{3}+(-1)^62^{4}
=-2+4-8+16=10$$
