Sequences and Series - Pre-Calculus

An arithmetic sequence is one in which there is a common difference between consecutive terms. For example, the sequence {2, 5, 8, 11} is an arithmetic sequence, because each term can be found by adding three to the term before it.

Let denote the nth term of the sequence. Then the following formula can be used for arithmetic sequences in general:

, where d is the common difference between two consecutive terms.

We are given the 4th and 8th terms in the sequence, so we can write the following equations:

.

We now have a system of two equations with two unknowns:

Let us solve this system by subtracting the equation from the equation . The result of this subtraction is

.

This means that d = 2.5.

Using the equation , we can find the first term of the sequence.

Ultimately, we are asked to find the hundredth term of the sequence.

The answer is 220.

The sum of all odd numbers is another way to construct perfect squares. To see why this is, we can construct the series as follows.

We draw from the series by subtracting one from each term.

We discard the 0 term and factor 2 out of the remaining terms.

And finally we use the property that to evaluate the series.

The smallest value of where the square exceeds 200 is .

The terms of an arithmetic series are generated by the relation

,

where is the 1st term, is the nth term, and d is the common difference.

For

,

for

.

The first step is to find .

, so

.

Now to find , when .

Use the generating relation

.

To find the next term, we need to figure out what is happening from one term to the next.

From 2 to 5, we can see that 3 is added.

From 5 to 14, 9 is added.

From 14 to 41, 27 is added.

If you look closely, you can notice a trend.

The amount added each time triples. Therefore, the next amount added should be

Thus,

This equation is a series in summation notation.

We can see that the bottom k=3 designates where the series starts, 8 represents the stop point, and 1/k represents the rule for summation. We can expand this equation as follows:

Here, we have just substituted "k" for each value from 3 to 8. To solve, we must then find the least common demoninator. That would be 280. This can be found in several ways, such as separating the fractions by like denominators:

In the series given, is added to each previous term to get the next term. Since a fixed number is ADDED each time, this series can be categorized as an arithmetic series.

First, we need to figure out what the pattern is in this series. Notice how each term results in the following term. In this case, each term is multiplied by to get the next term. Since each term is MULTIPLIED by a fixed number, this can be defined as a geometric series.

Common ratio is the number that is multiplied by each term to get the next term in a geometric series. Since the first two terms are and , we look at what is multiplied between these. Once way to determine this if not immediately obvious is to divide the second term by the first term. In this case we get:

which gives us our common ratio.

First, let's find a pattern for this sum. Each value has a difference of 3. If we know that the first value is 8, and that k will start at 1, and that each value must go up by 3, we can write the following:

Having determined the the rule for this sum, we can now determine what value it must end at by setting the rule function equal to the last value, 26.

Thus the summation notation can be expressed as follows:

We note that there is no common difference between and so the sequence cannot be arithmetic.

We also note that there exists a common ratio between two consecutive terms.

Since there exists a common ratio, the sequence is Geometric.

is defined as the sum of the terms from to

Therefore, to get the solution we must add all the entries from from to as follows.

Since the upper bound of the iterator is and the initial value is , we need add one-half, the summand, six times.

This results in the following arithmetic.

Because the iterator starts at , we first have a .

Now expanding the summation to show the step by step process involved in answering the question we get,

The answer is . Try this for :

This can be proven more generally using a proof technique called mathematical induction, which you will most likely not learn in high school.

Partial sums (written ) are the first few terms of a sum, so

If you then just take off the last number in that sum you get the and so on.

First break down the decimal into a sum of fractions to see the pattern.

and so on. Thus,

These fractions can be reduced, and the sum becomes

Each term is multiplied by to get the next term which is added.

For the first 4 terms this would look like

Let be the index variable in the sum, so if starts at the terms in the above sum would look like:

.

The decimal is repeating, so the pattern of addition occurs an infinite number of times. The sum expressed in sigma notation would then be:

.

In order to evaluate the summation, we must understand what the notation of the expression means:

This sigma notation tells us to sum the values obatined from evaluating the expression at each integer between and including those below and above the sigma. So we're going to start by evaluating the expression at n=1, and then add the value of the expression evaluated at n=2, and so on, until we end by adding the last value of the expression evaluated at n=5. This process is shown mathematically below:

The summation starts at 6 and ends at 7. Increase the value of after each iteration:

Rewrite the summation term by term:

To simplfy we get a common denominator of 24.