Answer
a) $1+2n^2$
b) $c$
c) $\dfrac{n+1}{2n}$
Work Step by Step
Use formula: $\Sigma_{k=1}^n k=\dfrac{n(n+1)}{2}$ and $\Sigma_{k=1}^n k^2=\dfrac{n(n+1)(n+2)}{6}$
a) Here, we have $\Sigma_{k=1}^{n} (\dfrac{1}{n}+2n)=(\dfrac{1}{n}+2n)\Sigma_{k=1}^{n}(1)=(\dfrac{1}{n}+2n)(n)=1+2n^2$
b) Here, we have $\Sigma_{k=1}^{n} \dfrac{c}{n}=(\dfrac{c}{n})(n)=c$
c) Here, we have $\Sigma_{k=1}^{n} \dfrac{k}{n^2}=(\dfrac{1}{n^2})(\dfrac{n(n+1)}{2})=\dfrac{n+1}{2n}$
