How do i simplify the following: $\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)$ I want to simplify

$$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)$$

without a summation sign in the answer. How can I do this?
I can't work out how to simplify this expression. I appreciate any help.
 A: Well the obvious first step will to try to see if we can combine terms:
$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)=$
$(3\cdot1^2 + 4) + \sum_{k=2}^n (3k^2+4) - \sum_{k=2}^n (3k^2+1) - [3(n+1)^2 + 1]=$
$(3\cdot1^2 + 4)+\sum_{k=2}^n[(3k^2+4)- (3k^2+1)] - [3(n+1)^2 + 1]$
and see what happens.  As it turns out a lot happens.  (All of it good.)
A: Hint: Sometimes it helps to write the terms of a sum so you're not just looking at symbols. The sum is
\begin{align*}
(3 \cdot 1^2 + 4) + &(3 \cdot 2^2 + 4) + (3 \cdot 3^2 + 4) + \cdots + (3 \cdot n^2 + 4) \\
- \Big[ &(3 \cdot 2^2 + 1) + (3 \cdot 3^2 + 1) + \cdots + (3 \cdot n^2 + 1) + (3 \cdot (n+1)^2 + 1)\Big].
\end{align*}
Can you see how to simplify that?
A: That solves it. Expand the terms .....
  
A: $$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{j=1}^{n} (3(j+1)^2+1)$$
$$=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} (3(i+1)^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} 3i^2+6i+4$$
$$=\sum_{i=1}^{n} -6i=-6\frac{n}{2}(n+1)=-3n(n+1)$$
