Change a summation range with two partial sums

Let $F(m)=m(m+1)(2m+7)/6$. The required sum is $F(2n)-F(n-1)$, retaining the term at $r=n$.

$F(2n)-F(n-1)=\dfrac n6[2(2n+1)(4n+7)-(n-1)(2n+5)]=\dfrac n6(14n^2+33n+19)=\dfrac{n(n+1)(14n+19)}6$.