sequences and series - Summation of $1cdot 3cdot 5cdot 7 + 3cdot 5cdot 7cdot 9 ...$

Find the sum of:

$1 \cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9+...$ till $n$ terms.

My attempt:

I got the $i^{th}$ term to be $(2i-1)(2i+1)(2i+3)(2i+5)$

Expansion gives: $16i^4 +64i^3+56i^2+-16i-15$

Required: $$\sum\limits_{i=1}^n (16i^4 +64i^3+56i^2+-16i-15) $$

Using summation identities, I got:

$\dfrac{16n(n+1)(2n+1)(3n^2+3n-1)}{30}+\dfrac{64n^2(n+1)^2}{4}+\dfrac{56(n)(n+1)(2n+1)}{6}- \dfrac{16n(n+1)}{2}- 15n$

However, answer given is simply $$\frac{1}{10}\{(2n-1)(2n+1)(2n+3)(2n+5)(2n+7)+1\cdot 3\cdot 5\cdot 7\}$$