Answer
a) $4n$
b) $cn$
c) $\dfrac{n^2-n}{2}$
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 $\Sigma_{k=1}^{n} (4)=4n$
b) Here, we have $\Sigma_{k=1}^{n} (c)=cn$
c) Here, we have $\Sigma_{k=1}^{n} (k-1)=\Sigma_{k=1}^{n} (k)-\Sigma_{k=1}^{n} (1)=\dfrac{n^2-n}{2}$