Question Video: Evaluating the Sum of a Finite Quadratic Series with a Starting Index Greater Than 1 Using the Properties of Summation Mathematics

Video Transcript

Given that the sum from 𝑟 equals one to 𝑛 of 𝑟 is equal to 𝑛 multiplied by 𝑛 plus one over two and the sum from 𝑟 equals one to 𝑛 of 𝑟 squared is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one over six, use the properties of the summation notation 𝛴 to find the sum from 𝑟 equals five to eight of five 𝑟 squared minus 22𝑟.

In this problem, we are asked to evaluate a quadratic series with an index 𝑟 ranging from five to eight. To do this, we’ll need to make use of a number of properties of summation. As the starting index 𝑟 is five, we first recall the property for starting indices greater than one. The sum from 𝑟 equals 𝑚 to 𝑛 of 𝑎 sub 𝑟 is equal to the sum from 𝑟 equals one to 𝑛 of 𝑎 sub 𝑟 minus the sum from 𝑟 equals one to 𝑚 minus one of 𝑎 sub 𝑟. Applying this property, we have that the sum from 𝑟 equals five to eight of five 𝑟 squared minus 22𝑟 is equal to the sum from 𝑟 equals one to eight of five 𝑟 squared minus 22𝑟 minus the sum from 𝑟 equals one to four of five 𝑟 squared minus 22𝑟.

Next, as the summand is the sum of two terms, we also recall the summation linearity property. For constants 𝜆 one and 𝜆 two, the sum from 𝑟 equals one to 𝑛 of 𝜆 one multiplied by 𝑎 sub 𝑟 plus 𝜆 two multiplied by 𝑏 sub 𝑟 is equal to 𝜆 one multiplied by the sum from 𝑟 equals one to 𝑛 of 𝑎 sub 𝑟 plus 𝜆 two multiplied by the sum from 𝑟 equals one to 𝑛 of 𝑏 sub 𝑟. In other words, we can separate the sum into two separate sums and each time bring the constants out the front.

Applying this property to our new version of the sum, we have that the sum from 𝑟 equals five to eight of five 𝑟 squared minus 22𝑟 is equal to five multiplied by the sum from 𝑟 equals one to eight of 𝑟 squared minus 22 multiplied by the sum from 𝑟 equals one to eight of 𝑟. And from this, we’re subtracting five multiplied by the sum from 𝑟 equals one to four of 𝑟 squared minus 22 multiplied by the sum from 𝑟 equals one to four of 𝑟.

At this point, we need to recall the standard results for the sum from 𝑟 equals one to 𝑛 of 𝑟 and the sum from 𝑟 equals one to 𝑛 of 𝑟 squared, which are each given in the question. Substituting 𝑛 equals eight, the first two terms become five multiplied by eight multiplied by eight plus one multiplied by two times eight plus one over six minus 22 multiplied by eight multiplied by eight plus one over two. And then, we substitute 𝑛 equals four to evaluate the second two terms. Evaluating gives 1020 minus 792 minus 150 plus 220, which is 298.

So, by recalling the property for summing a series with a starting index greater than one and the summation linearity property together with the standard results for the sum from 𝑟 equals one to 𝑛 of 𝑟 and the sum from 𝑟 equals one to 𝑛 of 𝑟 squared, we found that the sum from 𝑟 equals five to eight of five 𝑟 squared minus 22𝑟 is 298.