Problem 31 - Precalculus: Mathematical for Calculus

Question

Find the first six partial sums \(S_{1}, S_{2}, S_{3}, S_{4}, S_{5}, S_{6}\) of the sequence. $$1,3,5,7, \dots$$

Step-by-Step Solution

Answer

The first six partial sums are 1, 4, 9, 16, 25, and 36.

Step 1: Understand the Sequence

The sequence given is an arithmetic sequence starting with 1 and increasing by 2 each time. This can be represented as 1, 3, 5, 7, 9, 11, etc. The general term for the sequence is given by \(a_n = 2n - 1\).

Step 2: Calculate the First Term (\(S_1\))

The first partial sum is just the first term of the sequence. Thus, \(S_1 = 1\).

Step 3: Calculate the Second Partial Sum (\(S_2\))

The second partial sum is the sum of the first two terms of the sequence: \[ S_2 = 1 + 3 = 4. \]

Step 4: Calculate the Third Partial Sum (\(S_3\))

Add the third term of the sequence to \(S_2\): \[ S_3 = S_2 + 5 = 4 + 5 = 9. \]

Step 5: Calculate the Fourth Partial Sum (\(S_4\))

Sum the first four terms of the sequence: \[ S_4 = S_3 + 7 = 9 + 7 = 16. \]

Step 6: Calculate the Fifth Partial Sum (\(S_5\))

Sum the first five terms of the sequence: \[ S_5 = S_4 + 9 = 16 + 9 = 25. \]

Step 7: Calculate the Sixth Partial Sum (\(S_6\))

Add the sixth term to get the sixth partial sum: \[ S_6 = S_5 + 11 = 25 + 11 = 36. \]

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a series of numbers where each term after the first is generated by adding a constant difference to the previous term. This type of sequence is easy to spot due to its regularity.
In an arithmetic sequence, every step involves the same amount of change. For example, in our exercise with the sequence 1, 3, 5, 7, ..., the constant difference is 2. This means we add 2 to each term to get the next one.

• The first term is 1.
• The second term is 3, calculated as 1 + 2.
• The third term is 5, which is 3 + 2.

Recognizing these patterns is crucial because it allows us to predict any term in the sequence. We don’t have to perform the additions repeatedly; instead, we can use a general term formula to find any term.

General Term

A general term of an arithmetic sequence is a formula that allows you to find any term in the sequence without having to list out all the previous terms.
This formula is particularly useful when calculating terms far into the sequence.
The general term of an arithmetic sequence can be expressed as:

• \( a_n = a_1 + (n-1) imes d \)

where \( a_n \) is the nth term, \( a_1 \) is the first term, \( n \) is the number of the term, and \( d \) is the common difference between consecutive terms.
In our specific case, the formula becomes:

• \( a_n = 2n - 1 \)

This tells us the nth term of the sequence. For example, to find the fifth term (\( a_5 \)), you replace \( n \) with 5 in the equation, giving \( a_5 = 2 \times 5 - 1 = 9 \).
So you can quickly find any term without listing each one.

Sequence

A sequence is an ordered list of numbers that follow a specific pattern. Sequences can be arithmetic, like the one in our example, geometric, or more complex.
Understanding sequences involves recognizing these patterns and employing formulas to find terms efficiently. For this exercise, we dealt with an arithmetic sequence, but sequences can take many forms. Some may multiply by a constant factor (geometric) or follow even more complex rules.
In any sequence, terms are usually denoted by \( a_1, a_2, a_3, \ldots \). The position of the term in the sequence gives it the subscript \( n \) in the general term formula.
Mastering sequences includes not just identifying the pattern but using those patterns to make calculations simpler and faster, like finding partial sums without needing to track each and every step.
This understanding aids in solving various mathematical problems efficiently.