Problem 56 - Algebra and Trigonometry

Question

Expand each sum. \(\sum_{k=0}^{n-1}(2 k+1)\)

Step-by-Step Solution

Answer

\( \sum_{k=0}^{n-1}(2k+1) = n^2\).

Step 1: Understanding the Summation Notation

The summation notation \(\sum_{k=0}^{n-1}(2k+1)\) means that we need to compute the sum of the expression \(2k+1\) as k takes on integer values from 0 to \(n-1\).

Step 2: Write Out the Terms

Let's write out the terms for the given summation when k takes values from 0 to \(n-1\): \( (2\cdot0+1) + (2\cdot1+1) + (2\cdot2+1) + \ldots + (2(n-1)+1) \) This simplifies to: \( 1 + 3 + 5 + \ldots + (2n-3) + (2n-1) \)

Step 3: Calculate the Sum of the Sequence

The sequence \(1, 3, 5, \ldots, (2n-1)\) is an arithmetic series with a common difference of 2. The general formula for the sum of an arithmetic series is \[ S = \frac{n}{2} [2a + (n-1)d] \] where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms. Here, \(a=1\), \(d=2\), and there are \(n\) terms.

Step 4: Apply the Formula

Using the formula for the sum of the arithmetic series: \[ S = \frac{n}{2} [2\cdot1 + (n-1)\cdot2] \] Simplify inside the brackets: \[ S = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} [2n] = n^2 \] Thus, the sum \(\sum_{k=0}^{n-1}(2k+1)=n^2\).

Step 5: Final Expression

The expanded sum of \( \sum_{k=0}^{n-1}(2k+1) \) is \(n^2\).

Key Concepts

arithmetic series

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This is called the common difference. For example, in the sequence 1, 3, 5, 7, the common difference is 2.

When dealing with an arithmetic series, you can find the sum of all terms using a specific formula.

Consider the following sequence: 1, 3, 5, \ldots, (2n-1). This series has a first term (a) of 1 and a common difference (d) of 2. To find the sum (S) of the first n terms, we use the formula:

• \[ S = \frac{n}{2} [2a + (n-1)d] \]

In this context:

• a = 1
• d = 2
• n = number of terms

By plugging these values into the formula, you can find the sum for any number of terms in the series.

sum of integers

The sum of integers in a sequence can be calculated easily if you know the type of sequence. For arithmetic series, where each term has a constant difference, the formula to find the sum helps a lot.

Consider the sequence of integers: 1, 3, 5, ..., (2n-1). Each of these numbers is formed by starting at 1 and then adding 2 successively.

• This sequence continues up to (2n-1).
• Note that the number of terms (n) will be important in our formula.

Using the arithmetic series sum formula, we previously found:

• The sum S = \[ \frac{n}{2} \[2 \cdot 1 + (n-1) \cdot 2\] \]

By simplifying the expression inside the brackets, we get:\[ \frac{n}{2} [2 + 2n - 2] = n \cdot n = n^2 \]Thus, the sum of integers in this specific arithmetic series is \( n^2 \). This specific formula can help you quickly compute sums without adding each integer individually.

nth term

The nth term of a sequence provides a formula to find any term given its position in the sequence. For our arithmetic series, the nth term can be useful for understanding the entire sequence.

In the series 1, 3, 5, \ldots, (2n-1), each term gets its value from the formula for the nth term of an arithmetic sequence, which is:

• nth term = a + (n-1)d

In this case:

• a = 1 (the first term)
• d = 2 (the common difference)

So, the nth term calculation becomes:

• (2n-1) = 1 + (n-1)2

Hence, if you want to find the value of any term in this sequence, just plug in the value of n in the above formula.
For example:

• The 1st term (n=1) is 1 \(1 = 1+(1-1)2\)
• The 2nd term (n=2) is 3 \(3 = 1+(2-1)2\)
• The 3rd term (n=3) is 5 \(5 = 1+(3-1)2\)

This helps to identify patterns and quickly spot-check any term in your sequence.