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

$3376$
Use formula: $\Sigma_{k=1}^n k=\dfrac{n(n+1)}{2}$ and $\Sigma_{k=1}^n k^2=\dfrac{n(n+1)(n+2)}{6}$ Here, we have $\Sigma_{k=1}^{5} \dfrac{k^3}{225}+(\Sigma_{k=1}^{5} k)^2=\dfrac{1}{225}[\Sigma_{k=1}^{5} k^3]+(\dfrac{5 \cdot 6}{2})^3=1+3375=3376$