Question

Find the sum of each sequence. \(\sum_{k=0}^{14}\left(k^{2}-4\right)\)

Short answer

The sum is 955.

Step-by-step solution

Understand the Summation Notation

The sum can be written as \[\sum_{k=0}^{14}(k^{2} - 4)\].This notation means that you need to calculate the sum of the expression \((k^{2} - 4)\) as k goes from 0 to 14.

Separate the Summation

Use the linearity of summation to separate the sum into two parts:\[\sum_{k=0}^{14}(k^{2} - 4) = \sum_{k=0}^{14}k^{2} - \sum_{k=0}^{14}4.\]You will calculate each part separately.

Sum of \(k^{2}\)

To find the sum \(\sum_{k=0}^{14}k^2\):Use the formula for the sum of squares: \(\sum_{k=0}^{n-1}k^2 = \frac{(n-1)n(2n-1)}{6}\). Here, \(n = 15\).\[\sum_{k=0}^{14}k^2 = \frac{(14)(15)(29)}{6} = 1015.\]

Sum of Constants

Calculate the sum of the constant term:\[\sum_{k=0}^{14}4 = 4\times 15 = 60.\]

Combine the Results

Subtract the sum of the constant term from the sum of squares term:\[1015 - 60 = 955.\]

Key concepts

Sum of Squares

In mathematics, the 'sum of squares' is the sum of the squares of a series of numbers. This is often used when computing the sum of the square of each number in a sequence. The formula to find the sum of squares up to a number is given by: \[ \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \].

This formula allows you to find the sum without having to square and then add each number manually. For example, if we want to find the sum of squares from 0 to 14, we would set n to 15, since we're summing up to 14 (-1). Plugging in the values, you get: \[ \sum_{k=0}^{14} k^2 = \frac{(14)(15)(29)}{6} \].
Breaking it down step-by-step:

• (14 \times 15 \times 29) = 6090
• Then, you divide 6090 by 6 to get 1015

So, the sum of the squares from 0 to 14 is 1015.

Linearity of Summation

The linearity of summation is a property that allows us to break down a sum into separate parts. This means if you have a complex sum involving addition or subtraction inside, you can split it into simpler sums. For instance, if you have \sum_{k=0}^{14}(k^{2} - 4) , you can use linearity to separate it: \[ \sum_{k=0}^{14}(k^{2} - 4) = \sum_{k=0}^{14} k^{2} - \sum_{k=0}^{14} 4 \].
Basically, you are taking the entire expression and breaking it down into two simpler parts. This makes it easier to handle and compute sums. Once separated, compute each part and then combine the results. For our example:

• \sum_{k=0}^{14} k^2 = 1015
• \sum_{k=0}^{14} 4 = 60

Now, subtract 60 from 1015, which results in 955.

Constant Term Summation

Summing a constant term is straightforward. If you are adding or subtracting a constant term repeatedly over a range in a summation notation, you can simplify the process. For example, \sum (k=0)^{14} 4 means you add the number 4 fifteen times since you start from k=0 and go up to k=14. To find the sum, you multiply 4 by the number of terms. The number of terms is determined by the values of k, in this case: k=0 to k=14 constitutes 15 terms.
So, \[ \sum(k=0)^{14} 4 = 4 \times 15 = 60 \].
Understanding this concept is fundamental because it makes solving the part of the summation involving constants very simple and quick. Once you know how to sum a constant term, you apply the result to either add or subtract from other parts of the summation.