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Evaluating a sum by splitting into simpler sums

Consider the finite sum ∑i=18(i2−3i+2)\displaystyle\sum_{i=1}^{8}(i^2 - 3i + 2).

Evaluate ∑i=18(i2−3i+2)\sum_{i=1}^{8}(i^2 - 3i + 2) by separating it into three sums involving ∑i2\sum i^2, ∑i\sum i, and ∑1\sum 1, and using the closed-form formulas ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2} and ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}. What is the numerical value of the sum?

Hint

Use linearity to write the sum as ∑i2−3∑i+∑2\sum i^2 - 3\sum i + \sum 2, then apply the closed-form formulas with n=8n=8 (remembering ∑i=182=2⋅8\sum_{i=1}^{8} 2 = 2\cdot 8).

Step-by-step solution

By linearity, ∑i=18(i2−3i+2)=∑i=18i2−3∑i=18i+∑i=182\sum_{i=1}^{8}(i^2-3i+2) = \sum_{i=1}^{8} i^2 - 3\sum_{i=1}^{8} i + \sum_{i=1}^{8} 2. Using the formulas with n=8n=8: ∑i=18i2=8(9)(17)6=204\sum_{i=1}^{8} i^2 = \frac{8(9)(17)}{6} = 204, and ∑i=18i=8(9)2=36\sum_{i=1}^{8} i = \frac{8(9)}{2} = 36. Also ∑i=182=2(8)=16\sum_{i=1}^{8} 2 = 2(8) = 16.

Substituting these values gives 204−3(36)+16=204−108+16=112204 - 3(36) + 16 = 204 - 108 + 16 = 112. So the sum evaluates to 112112.

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