Sum of Cubes

k(k² + n) = (k³ + nk)

k³ is in the sigma.

See ∑ (+nk).
k is the variable of the sigma.
So think n as a coefficient.
Then n gets out from the sigma.
And k is in the sigma.

Summation: How to Solve

The sum of k³ is
[n(n + 1)/2]².

Write the 'coefficient' +n.

The sum of k is
n(n + 1)/2.

Sum of k

So [n(n + 1)/2]² + n⋅n(n + 1)/2.

[n(n + 1)/2]²
= n²(n + 1)²/2²
= n²(n + 1)²/4

Power of a Product

Power of a Quotient

+n⋅n(n + 1)/2 = +n²(n + 1)/2

To combine these two fractions,
change +n²(n + 1)/2 to +2n²(n + 1)/4.

The common factor of
n²(n + 1)²/4 and +2n²(n + 1)/4
is n²(n + 1)/4.

So write
n²(n + 1)/4.

Write, n²(n + 1)²/4 ÷ n²(n + 1)/4,
[(n + 1).

And write, +2n²(n + 1)/4 ÷ n²(n + 1)/4,
+2].

[(n + 1) + 2] = (n + 3)

So
n(n + 1)(n + 3)/4
is the answer.