We started of our lecture with some "quickies"...

1. Find the value(s) of "r" in 5r^4 = 80

Solution:

In order to solve this question, we just use algebraic methods...

2. In the geometric sequence, if t1 = 3 and r = 2, find t3.

Solution:

If we can recall the rule for geometric sequences from last class...3. If the first term of a geometric progression is 1/3 and the common ratio is -3, find the next 3 terms.

Solution:

In order to solve this, we can simply use the geometric sequence rule to find t2, t3, and t4...

4. Determine the common ratio for the geometric sequence; 1/(x^1/2), 1, (x^1/2).

Solution:

In order to solve this problem, first we pick either the first, second, or third term. We can concentrate on that term to find the common ratio, "r". There is one catch, if we let n = 1, our common ratio will be 1 since "n-1" is the exponent of the ratio. If we use n = 1, then the exponent will be 0, meaning, r will equal 1 (anything to the power of 0 is 1)...

I'll pick the 2nd term...Then, Mr. K told us the story of the great

**Johann Carl Friedrich Gauss**

So, I'll try to tell the story.. XD

One day, Gauss' teacher asked his students to "find the sum of the numbers from 1 to 100". During his time, there were no papers to write on, just slates. His classmates started to add up all the numbers, starting 1. (Man... this is gonna take forever...) Gauss didn't touch his slate and thought about it for approximately 1.5 to 2 mins. He wrote the number 5050 on his slate and handed it in to his teacher. (Even the teacher doesn't know the answer to this one...)

The big question was... "How in the world did he do that?!"

This is how he did it...

1 + 2 + 3 + 4 .... 97 + 98 + 99 + 100

100 + 99 + 98 + 97.... 4 + 3 + 2 + 1

101 101 101 101 .... 101 101 101 101

If we add 2 copies of the sequence, in which one is written in reverse order, we could see a pattern...

Then, Gauss simply put together these numbers...It means that the number 101 has been obtained 100 times in his method. He figured that this answer has been added twice. Therefore, he multiplied the answer by 1/2.

In other words...

t | + | (t + d) | + | (t + 2d) | + ... + | (t + nd) |

(t + (n-1)d) | + | (t + (n-2)d) | + | (t + (n-3)d) | + ... + | t |

(2x + (n-1)d) | + | (2x + (n-1)d) | + | (2x + (n-1)d) | + ... + | (2x + (n-1)d) |

Since 2t + nd = t + (t + nd) = t

_{0}+ t

_{n},

For any arithmetic sequence, the series could be found by using the Arithmetic Summation...

x+ (x + d) + (x + 2d) + ... + (x + nd) = (n + 1)(a

_{0}+ a

_{n})/2

or

Where:

"Sn" is the sum of all the numbers up to the "nth" term

" n " is the rank of the nth term

" t1" is the first term in the sequence

" d " is the common difference

A series is the sum of numbers in a sequence to a particular term in a sequence.

For a geometric sequence...

Sn = t1 ( 1 - r^n)/(1 - r)

Where:

"t1" is the first term

"r" is the common ratio

"n" is the rank of the term

Finally, we talked about the "Sigma Notation". It's a short-hand way of writing a series.

The "funky" looking "E" is a Greek letter for "sum". The number above the "E" means "to find the value of n, until we have obtained the nth term of the sequence. "n-1" means that this is the initial value of the sequence; meaning, you have to start evaluating at n-1. "2n - 1" is the "rule" of the given sequence.

That's what we did for this class. (Finally, I have edited my scribe post! Yay!)

Oh well, good night everyone! I'm tired...

Next scribe will be... once again... "BERTMAN" XD.

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