Question Video: Evaluating the Sum of a Finite Quadratic Series Using the Properties of Summation Mathematics

Given βˆ‘_(π‘Ÿ = 1) ^(𝑛) π‘ŸΒ² = (𝑛(𝑛 + 1)(2𝑛 + 1))/6, use the properties of summation to find βˆ‘_(π‘Ÿ = 1) ^(6) (5π‘ŸΒ² βˆ’ 67).

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Video Transcript

Given the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one over six, use the properties of summation to find the sum from π‘Ÿ equals one to six of five π‘Ÿ squared minus 67.

In this question, we’re asked to find the sum of a quadratic series in which the summand is five π‘Ÿ squared minus 67. And we’re summing from π‘Ÿ equals one to π‘Ÿ equals six. As the summand is the sum or in fact the difference of two terms, we can recall the summation linearity property. This states that for constants πœ† one and πœ† two, the sum from π‘Ÿ equals one to 𝑛 of πœ† one π‘Ž π‘Ÿ plus πœ† two 𝑏 π‘Ÿ is equal to πœ† one multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ž π‘Ÿ plus πœ† two multiplied by the sum from π‘Ÿ equals one to 𝑛 of 𝑏 π‘Ÿ.

Applying this result then with πœ† one equal to five, π‘Ž π‘Ÿ equal to π‘Ÿ squared, πœ† two equal to negative one, and 𝑏 π‘Ÿ equal to 67, we have that the sum from π‘Ÿ equals one to six of five π‘Ÿ squared minus 67 is equal to five multiplied by the sum from π‘Ÿ equals one to six of π‘Ÿ squared minus the sum from π‘Ÿ equals one to six of 67. We’ll be able to evaluate the first of these sums using the general result stated in the question. But to evaluate the second, we need to recall a second general result. Here, we’re summing the constant 67. So we recall that the sum from π‘Ÿ equals one to 𝑛 of a constant 𝛼 is equal to 𝛼 multiplied by 𝑛. So let’s go ahead and evaluate these two sums.

Substituting 𝑛 equals six into the general result given in the question for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared, we have five multiplied by six multiplied by six plus one multiplied by two times six plus one all over six. Then, applying the standard result for the sum from π‘Ÿ equals one to 𝑛 of a constant with 𝛼 equal to 67 and 𝑛 equal to six, we are subtracting 67 multiplied by six. Simplifying, we have five multiplied by six multiplied by seven multiplied by 13 over six minus 402. We can then cancel a factor of six in the numerator and denominator of the first term. And we’re left with five multiplied by seven multiplied by 13, which is 455 minus 402, which is equal to 53.

So, using the summation linearity property together with the standard results for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared and the sum from π‘Ÿ equals one to 𝑛 of a constant, we found that the sum from π‘Ÿ equals one to six of five π‘Ÿ squared minus 67 is 53.

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