Summation of $1\cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9 ...$ 
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 $$\frac{1}{10}\{(2n-1)(2n+1)(2n+3)(2n+5)(2n+7)+1\cdot 3\cdot 5\cdot 7\}$$

Let
$$
f(k) = (2k-1) \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
.
$$
Then
$$
f(k+1) = \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}(2k+9)
,
$$
so
$$
f(k+1)-f(k) = 10 \, \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
,
$$
and your sum can be rewritten as the telescoping sum
$$
\frac{1}{10}\sum_{k=0}^{n-1} \Bigl( f(k+1)-f(k) \Bigr)
= \frac{1}{10} \Bigl( f(n)-f(0) \Bigr)
= \frac{f(n)- (-1)\cdot 1 \cdot 3 \cdot 5 \cdot 7}{10}
.
$$
(Which agrees with your answer, by the way, if you expand everything out and compare.)
