TITLE: How do i simplify the following: $\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)$
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: $$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{j=1}^{n} (3(j+1)^2+1)$$
$$=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} (3(i+1)^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} 3i^2+6i+4$$
$$=\sum_{i=1}^{n} -6i=-6\frac{n}{2}(n+1)=-3n(n+1)$$
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