Find the sum.
-$\sum _ { k = 1 } ^ { n } \left( k ^ { 3 } + k ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The sum $\sum _ { k = 1 } ^ { n } \left( k ^ { 3 } + k ^ { 2 } \right)$ can be separated into two sums: $\sum _ { k = 1 } ^ { n } k ^ { 3 }$ and $\sum _ { k = 1 } ^ { n } k ^ { 2 }$. The formula for the sum of the first $n$ cubes is $\left( \frac{n(n+1)}{2} \right)^2$ and the formula for the sum of the first $n$ squares is $\frac{n(n+1)(2n+1)}{6}$. Substituting these into the original equation gives $\left( \frac{n(n+1)}{2} \right)^2 + \frac{n(n+1)(2n+1)}{6}$, which simplifies to $\frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$, matching choice C.

Find the Sum ∑k=1n(k3+k2)\sum _ { k = 1 } ^ { n } \left( k ^ { 3 } + k ^ { 2 } \right)∑k=1n​(k3+k2)

Find the Sum $\sum _ { k = 1 } ^ { n } \left( k ^ { 3 } + k ^ { 2 } \right)$