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Evaluating a finite sum using summation formulas

Use the summation formulas ∑i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2} and ∑i=1ni3=(n(n+1)2)2\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 to evaluate the sum

∑i=1n(i3+2i)\sum_{i=1}^{n} (i^3 + 2i)

in terms of nn. Write your final answer as a simplified expression in nn.

Hint

Split the sum into two separate sums using linearity, then apply the known closed-form formula for ∑i3\sum i^3 and for ∑i\sum i, and simplify.

Step-by-step solution

By linearity of summation, ∑i=1n(i3+2i)=∑i=1ni3+2∑i=1ni\sum_{i=1}^n (i^3+2i) = \sum_{i=1}^n i^3 + 2\sum_{i=1}^n i. Substitute the known formulas: ∑i=1ni3=(n(n+1)2)2\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 and ∑i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2}.

This gives (n(n+1)2)2+2⋅n(n+1)2=n2(n+1)24+n(n+1)\left(\frac{n(n+1)}{2}\right)^2 + 2\cdot\frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{4} + n(n+1). Factoring out n(n+1)n(n+1) yields n(n+1)(n(n+1)4+1)=n(n+1)(n2+n+4)4n(n+1)\left(\frac{n(n+1)}{4} + 1\right) = \frac{n(n+1)(n^2+n+4)}{4}.

Common factor in the simplified sum expression

Hint
Step-by-step solution

The answer is n. When simplifying the sum, we factored the expression as n times open parenthesis n plus one close parenthesis times the quantity n squared plus n plus four, all over four. The two factors pulled out front were n and n plus one, so besides n plus one, the other common factor is simply n.

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