<s>[INST] Simplify the subtraction of two summations. $$\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) $$
I thought I would use the distributive rule but the question stipulates that my answer should not include a summation sign... 
 [/INST] Method 1:
$$\begin{align}
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =\sum_{i=1}^n (3i^2 +4) - \sum_{j=1}^n (3(j+1)^2 +1) \\
& = \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2+6j+4) \\
& = \sum_{i=1}^n \left[(3i^2 +4) - (3i^2+6i+4)\right] \\
& = \sum_{i=1}^n -6i \\
& = -6\cdot\frac{n(n+1)}2 \\
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\
\end{align}$$

Method 2:
$$\begin{align}
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & = 7+\sum_{i=2}^n (3i^2 +4) - \sum_{j=2}^n (3j^2 +1)-3(n+1)^2-1 \\
& =6-3(n+1)^2+\sum_{i=2}^n \left[(3i^2 +4) - (3i^2 +1)\right] \\
& =6-3(n+1)^2+\sum_{i=2}^n3 \\
& =6-3(n+1)^2+\underbrace{3+3+3+\dots+3}_{n-1} \\
& =6-3(n+1)^2+3(n-1) \\
& =-3n^2-3n \\
\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\
\end{align}$$
</s>