Problem 48 - Contemporary Precalculus

Question

Find the sum. $$\sum_{n=1}^{75}(-1)^{n}(3 n+1)$$

Step-by-Step Solution

Answer

Answer: \(\frac{-751}{2}\)

Step 1: Identify the sequence formula

The given sum notation is: $$\sum_{n=1}^{75}(-1)^{n}(3 n+1)$$ The sequence can be written as \((-1)^{n}(3n+1)\).

Step 2: Write out the first few terms

By substituting the first few integer values of \(n\) in the sequence, we can find the first few terms. For n=1: \((-1)^{1}(3(1)+1)=-4\) For n=2: \((-1)^{2}(3(2)+1)=7\) For n=3: \((-1)^{3}(3(3)+1)=-10\) For n=4: \((-1)^{4}(3(4)+1)=13\) So, the sequence looks like: \(-4,7,-10,13,\dots\)

Step 3: Break the sum into two separate sums

We can break the sum into two separate sums with only positive terms and negative terms. $$\sum_{n=1}^{75}(-1)^{n}(3 n+1) = \sum_{n=1}^{37}(-1)^{2n-1}(3(2n-1)+1) + \sum_{n=1}^{38}(-1)^{2n}(3(2n)+1)$$

Step 4: Simplify the separate sums

Let's simplify the two separate sums. $$\sum_{n=1}^{37}(-1)^{2n-1}(3(2n-1)+1) = \sum_{n=1}^{37}(3(2n-1)+1)$$ $$\sum_{n=1}^{38}(-1)^{2n}(3(2n)+1) = \sum_{n=1}^{38}(3(2n)+1)$$

Step 5: Apply the arithmetic sum formula

Use the arithmetic sum formula \(\frac{n}{2}(a_1 + a_n)\) for both sums. We can now rewrite the sums as follows: $$\sum_{n=1}^{37}(3(2n-1)+1) = \frac{37}{2}(-4 + (1+(75-1))\times 3)$$ $$\sum_{n=1}^{38}(3(2n)+1) = \frac{38}{2}(7 + (2+(75-1))\times 3)$$

Step 6: Calculate the sums

Now, calculate both sums and find their difference. First sum: \(\frac{37}{2}(-4 + 74\times 3) = \frac{37}{2}(218)\) Second sum: \(\frac{38}{2}(7 + 76\times 3) =\frac{38}{2}(235)\) Difference: \(\frac{37}{2}(218) -\frac{38}{2}(235) = \frac{-751}{2}\) So, the sum of the sequence is $$\frac{-751}{2}$$.

Key Concepts

Alternating Series

An alternating series is a special kind of sequence in which the signs of the terms alternate between positive and negative. This might look like a daunting concept initially, but it's simply a pattern where terms flip between addition and subtraction. In our case, the formula \((-1)^{n}(3n+1)\) creates such a series. This is because \((-1)^n\) toggles the sign:

• For \(n = 1\): \((-1)^1(-1)\), making the term negative.
• For \(n = 2\): \((-1)^2(1)\), making the term positive.

Thus, each term either flips sign compared to its predecessor. Realizing this can drastically simplify the understanding of the series.
One way to approach these series is by breaking them into two subtasks: treating all negative and all positive terms separately. This method, as demonstrated in the solution, helps manage the complexity due to change of signs.

Sum of Series

Calculating the sum of a series, like the one in the problem \( \sum_{n=1}^{75}(-1)^{n}(3n+1)\), often involves breaking it down into simpler parts.
Initially, identifying the sequence formula is crucial. Once the sequence or pattern is recognized, as shown when substituting for small \(n\), the problem becomes less daunting.
The original series is divided into two separate series: one with only positive terms and one with only negative terms.

• Positive sum: \( \sum_{n=1}^{38}(3(2n)+1)\)
• Negative sum: \( \sum_{n=1}^{37}(3(2n-1)+1)\)

After separation, each of these individual series can be simplified and computed using the arithmetic series formula. Understanding the property of the series, in terms of positive and negative parts, simplifies the sum calculation, allowing one to focus on regular arithmetic sequences.

Arithmetic Sequence Calculation

An arithmetic sequence is quite straightforward. It's a series of numbers with the same difference between consecutive terms. To calculate the sum of such sequences, an arithmetic sum formula is powerful: \[ S_n = \frac{n}{2}(a_1 + a_n) \]where \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term.

• For the positive term series: \( \sum_{n=1}^{38}(3(2n)+1)\), we determine the last term using the general sequence formula \(a_n = a_1 + (n-1)d\).
• For the negative term series: \( \sum_{n=1}^{37}(3(2n-1)+1)\), the same procedure applies.

In both cases, "\(a_1\)" and "\(a_n\)" need evaluating through substitution into sequence formulae. After finding these values, the next step is straightforward: plug them into the arithmetic sum formula to get the sum of each part. By calculating each series separately and then combining as necessary, sums of more complex sequences become manageable.