First, we can distribute the summation to each term inside the parentheses:
$$\sum_{i=1}^{4}3 i^{3}+\sum_{i=1}^{4}2 i-\sum_{i=1}^{4}4$$
Next, we can evaluate each summation separately:
$$3\sum_{i=1}^{4}i^{3}+2\sum_{i=1}^{4}i-4\sum_{i=1}^{4}1$$
Now, let's evaluate
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