Scribe

Today in class we began the subject on Geometric Series. Based on the method Mr.A gave us for working on arithmetic serieses, we developed a method for the geometric series.

Our POD 5/10 was:
Find the sumof the 1st 30 positive odd integers:
So,
The sequence would be: 1,3,5,7,9,11...tn
*This is an arithmetic sequence since the diffrence between any two consecutive terms is 2. (ex:5-3=2, 9-7=2 etc)
The series would be: 1+3+5+7+9+11+....tn
Now,
Since we are asked for the sum od the first 30 terms in this sequence,we have to find the 30th term.
To do this we have to use the arithmetic sequence formula. Which is: tn=t1+ d(n-1)
Now, plug in d=2, t1=1, and n=30:
t30=1+2(30-1)
t30=1+58
t30=59

Then,
To find the sum of the first 30 terms of the series. We use the arithmetic series formula, which is:

Sn= (n/2)(t1+tn)

Now,
Plug in n=30 (to find the sum of the first 30 terms), t1=1, and tn=59 (as when n=30):
S30=(30/2)(1+59)
S30=(15)(60)
S30=900

So the answer for the problem would be: 900

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GEOMETRIC SERIES:

After the POD, Mr.A gave us the following problem to solve:

"A ball is dropped from 100 feet down to the ground and bounces 10 times befor coming to rest. On each bounce, it bounces half the height before. How far did it travel (vertically)?"

This problem was the first problem given to us about geometric series. Since none of the students knew how to develop the problem, Mr.A gave us the equation for a geometric series. It goes as follows:

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

Then he explained us the mathematics for getting this equation; a proof.

First,
he showed us what a geometric series looks like; which is:
Sn=t1+t1r+t1(r^2)+t1(r^3)+t1(r^4)+...t1(r^(n-1))

But,
This formula gives us more terms than we can handle and makes the process for solving the problem more difficult.

So,
He showed us a mathematical trick foreliminating the terms we don't really need.

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*A long time ago, some mathematician figured out this technique to prove the formula is right. We have to multiply "r" on both sides of the equation, then use that formula and subtract it from the original geometric series formula.

By doing this,
We eliminated the terms which we didn't really need and left us with:
Sn-rSn=t1-(t1(r^n))

Then,
By factoring: Sn(1-r)=t1(1-(r^n)) . We finally got the Geometric Series Formula.
It is:
Sn=t1((1-(r^n)/(1-r))

Now that we know the formula, we can go back to our problem and work it out.

Looking back at our picture in bubbleshare, we can see that the first bounce from the ground rises 50ft and drops 50ft, for a total of 100ft traveled in this bounce.

Consider:
Each term of our series involves 1 bounce, each bounce includes a height rising plus a height dropping. (Refer to bubbleshare drawing)

So,
The first section of the diagram, which includes just the initial drop of 100 ft, does not fit our definition of a bounce. Therefore, we just count this drop as a starting point for our series. So, we would add this 100ft to the sum of the first n terms. So our equation would look like this:

Sn=t1((1-(r^n)/(1-r)) + 100ft

Whith this information we can get that:

• # of bounces= 10 = n
• ratio of each bounce= 1/2 =r
• 1st bounce= 100ft = t1

Now,
We can use this information to plug into our geometric series formula:

S10=100((1-(0.5^10)/(1-o.5)) + 100
S10=199.804+100
S10=299.8

So,
The answer to our problem is: S10=299.8
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PRACTICE PROBLEMS:

To practice using the same techniques as the problem above, Mr. A gave us two problems to solve:

1. "Find the sum of the 1st 10 terms of the following geometric series: 5,25,125..."

Since,
For the geometric series formula we need t1, n, and r and the problem already gave us t1=5 and n=10; we look for r.

r=tn/tn-1
r=25/5
r=5

Now,
Plug into the formula:

S10=5((1-(5^10)/(1-5))
S10=12,207,030

So,
The answer to the problem is: S10=12,207,030
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*This next problem we didn't finish in class*

2. "If t1=1, t2=-3; Find S20.
a) If the sequence is geometric.
b) If the sequence is arithmetic.

Our solution:
a.
r=-3/1
r=-3

S20= 1((1-(-3^10))/(1-(-3)))
S20= -871,696,100

b) d= -3-1 = -4

t20= -1+(-4*19)

t20= -75

S20= (20/2)(1-75)

S20=-65

S20= -65

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Remember: Its better to understand where a formula came from and why is it so than to memorize it. Its going to be hard when you have to many formulas stuck in your head!

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HOMEWORK: Pg. 489, #9-14, 17-20, 27, 29 and 35.

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