University Calculus: Early Transcendentals (3rd Edition)

a) $1+2n^2$ b) $c$ c) $\dfrac{n+1}{2n}$
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 $\Sigma_{k=1}^{n} \dfrac{k}{n^2}=(\dfrac{1}{n^2})(\dfrac{n(n+1)}{2})=\dfrac{n+1}{2n}$