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Evaluating a finite sum using sigma notation properties

Evaluate the sum ∑j=38(j2−j)\sum_{j=3}^{8} (j^2 - j) by splitting it into separate sums, then using the closed-form formulas ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2} and ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}. What is the value of the sum?

Hint

Use linearity to write ∑j=38(j2−j)=∑j=38j2−∑j=38j\sum_{j=3}^{8}(j^2-j) = \sum_{j=3}^{8} j^2 - \sum_{j=3}^{8} j. Then express each sum from j=3j=3 to 88 as a difference of sums starting at 11, i.e. ∑j=38=∑j=18−∑j=12\sum_{j=3}^{8} = \sum_{j=1}^{8} - \sum_{j=1}^{2}, and apply the closed-form formulas.

Step-by-step solution

By linearity, ∑j=38(j2−j)=∑j=38j2−∑j=38j\sum_{j=3}^{8}(j^2-j) = \sum_{j=3}^{8} j^2 - \sum_{j=3}^{8} j. To use the standard formulas that start at i=1i=1, rewrite each piece as ∑j=38j2=∑j=18j2−∑j=12j2\sum_{j=3}^{8} j^2 = \sum_{j=1}^{8} j^2 - \sum_{j=1}^{2} j^2 and ∑j=38j=∑j=18j−∑j=12j\sum_{j=3}^{8} j = \sum_{j=1}^{8} j - \sum_{j=1}^{2} j.

Using ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}: ∑j=18j2=8⋅9⋅176=204\sum_{j=1}^{8} j^2 = \frac{8\cdot 9\cdot 17}{6} = 204 and ∑j=12j2=1+4=5\sum_{j=1}^{2} j^2 = 1+4=5, so ∑j=38j2=199\sum_{j=3}^{8} j^2 = 199. Using ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}: ∑j=18j=36\sum_{j=1}^{8} j = 36 and ∑j=12j=3\sum_{j=1}^{2} j = 3, so ∑j=38j=33\sum_{j=3}^{8} j = 33. Therefore the answer is 199−33=166199 - 33 = 166.

As a check, direct computation gives (9−3)+(16−4)+(25−5)+(36−6)+(49−7)+(64−8)=6+12+20+30+42+56=166(9-3)+(16-4)+(25-5)+(36-6)+(49-7)+(64-8) = 6+12+20+30+42+56 = 166, confirming the result.

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