Sequences And Series¶

Geometric and arithmetic sequences¶

Source: Made them up.

Explanation:

This artifact demonstrates the usage of both geometric and arithmetic sequences.

The first example is geometric, and the second is arithmetic.

Artifact:

Find the first 3 terms and the 50th term of the sequence \(\{a_k\}\) in which \(a_k = k^2-k\).

\[a_1 = 1^2 - 1 = 0\]\[a_2 = 2^2 - 2 = 2\]\[a_3 = 3^2 - 3 = 6\]\[a_50 = 50^2 - 50 = 2450\]

Find the first 3 terms and the 100th term of the sequence \(\{a_k\}\) in which \(a_k = k + 4\).

\[a_1 = 1 + 4 = 5\]\[a_2 = 2 + 4 = 6\]\[a_3 = 3 + 4 = 7\]\[a_100 = 100 +4 = 104\]

Defining sequences explicitly and recursively¶

Source: Group Quiz 9.4 & 9.5

Group Work:

Matthew and I disagreed on whether to start the index numbers at 0 or 1 for the explicit functions.

He thought we should start at 1, and I thought we should start at 0 because it was easier.

He was right, but we worked together to compromise.

We had one of them start at 0, and the other start at 1.

Explanation:

This artifact demonstrates defining sequences explicitly and recursively.

The recursive formula works by multiplying the previous number in the series by 3, starting with -2.

The explicit formula works by using indices (n) starting at 1.

The exponent on the formula is n-1. Since the first index is 1, the -3 will be nulled because anything to the 0 power is 1.

The explicit formula is a bit nicer for humans because you don’t have to calculate the values recursively.

Artifact:

Find the explicit and recursive formulas that model -2, 6, -18, 54, -162...

Recursive:

\[\begin{split}A_1 = -2\\A_n = A_{n-1} * 3\end{split}\]

Explicit:

\[A_n = -2(-3)^{n-1}\]

Summations notation¶

Source: Notes June 06, 2012

Explanation:

This artifact demonstrates summations notation.

The slope of the explicit function is 7 because that is the rate of change in the series that is suggested by the data.

I plugged in (1, 2) because 2 is the first item in the series (1 is the first index value).

I set it equal to the last term of the series to solve for the last index number, because that goes ontop of the sigma.

I used the gaussian method to find the sum of the finite arithmetic series.

Artifact:

Express the [2, 9, 16, 23, ..., 107] in summation notation.

\[y = mx + b\]\[y = 7x + b\]\[2 = 7(1) + b\]\[b = -5\]\[y = 7x -5\]\[107 = 7x - 5\]\[x = 16\]\[({(2 + 107) \over 2}) * 16 = 872\]\[\displaystyle\sum_{x=1}^{16} {7x-5} = 872\]

Summing finite arithmetic and geometric sequences¶

Source: Section 9.5 Example 2 and Group Quiz 9.4 & 9.5

Group Work:

The second example in this artifact comes our most recent group quiz.

I remember that we were very systematic and efficient in this quiz.

I calculated the first value in the series while he calculated the last one.

We were like Batman and Robin.

Matthew was Robin.

Explanation:

This artifact demonstrates summing finite arithmetic and geometric sequences.

I used the formula \(a_1(1 - r^n) \over 1 -r\) to find the sum of this geometric series.

I was given \(a_1 \text{ and } a_n\). I just needed to find the \(r\) value by using basic algebra.

I found the average of the first and last values in the series and then multiplied that by the number of values in the sequence to get the sum.

Artifact:

Find the sum of the geometric series \(4, -{4 \over 3}, {4 \over 9}, -{4 \over 27}, ... , 4(-{1 \over 3})^{10}\)

\[A_1 = 4\]\[4r = -{4 \over 3}\]\[r = -{1 \over 3}\]\[A_n = 4(-{1 \over 3})^{10}\]\[n = 11\]\[{4(1 + {1 \over 3}^{11}) \over 1 + {1 \over 3}} \approx 3.000016935\]

Find the sum of the arithmetic series \(\displaystyle\sum_{k=1}^{4} {-6 + k}\).

\[A_1 = -6 + 1 = -5\]\[A_4 = -6 + 4 = -2\]\[({(-5 +(-2)) \over 2}) * 4 = -14\]

Summing infinite geometric sequences¶

Source: Notes

Explanation:

This artifact demonstrates summing infinite geometric sequences.

To find the sum, I used the formula for the sum of an infinite geometric series (\(A_1 \over 1 - r\)).

This formula only works for geometric series that converge (eventually reaches a limit, usually 0).

Artifact:

Find the sum of the infinite geometric sequence [32, 16, 8, 4, 2, 1...]:

\[\text{sum } = {32 \over 1 - {1 \over 2}} = 64\]\[\displaystyle\sum_{k=1}^{\infty} {32({1 \over 2})^{k-1}} = 64\]