Scribe Post (May 10)

For those who did not understood, did not copy the notes or simply did not come to school, I hope this will help you.

Let me start of by saying that today there is a new homework, pg: 489 9-14, 17-20, 27, 29, 35.

As always, today we had a POD, that for some of us was a little bit tricky.

POD
Find the sum of the 1st 30 positive odd integers.1, 3, 5, ..., t30
*note: 30 is not the last number in the sequence!
Alex was the first one to find a correct solution. His solution was:
first find "d" (difference between jumps)
d= 2
Then by using the arithmetic sequences formula we just plug in the numbers
tn= t1+ (n-1) dt30= 1+ (30-1) 2= 1+ (29) 2= 1+ 58= 59
Then we aply the Arithmetic Series formula to find our answer.
S30= n/2 (t1+ t30)
= 30/2 (1+ 59)
= 15 (60)
= 900
900 is equal to the sum of all 1st 30 positive odd integers.
THANKS ALEX!

After the POD, Mr. A gave us a new challenge.

1.) A ball is dropped from 100 ft to the ground below where it bounces 10 times before comming to rest. On each bounce, the ball bounces to 1/2 of the height from which it just fell. How far did the ball travel?

How we did not have a formula to solve this, we all started adding up the numbers in the sequence.To get the sequence most of us did the following:
100/2= 50/2= 25/2=12.5, and so until we reached the 10th term.
Then add up 100, 50, 25, ... t10. If you did this, you should have got 199.8 ft (rounded). I'm sorry to say that the answer is not corre(ct.
(Picture1)
So, by adding all these numbers wickris' group got an answer of 295.78 ft (rounded), while jbarrika got and answer of 299.8 ft (rounded).
But to get to the real answer is not necesarry to divide everytime and then add because this is a GEOMETRIC SERIES.

A GEOMETRIC SERIES is a sum of the terms of a Geometric Sequence.
100+ sum of tn where tn= 1oo(1/2)^(n-1)
*note: these numbers are only for this problem!
We need a way to quickly get the sum of a geometric series.
[Sn= t1+ t1(r)+ t1(r^2)+ t1 (r^3)+ .... + t1(r^(n-1))]
However an easier way to get the answer is to now multiply r in both sides.
[rSn= t1(r)+ t1(r^2)+ t1 (r^3)+ .... + t1(r^(n-1))+ t1(r^n)]
(Picture 2)

So, going back into our problem...
S10= 100((1- .5^10)/(1- .5))= 199.8 + 100= 299.8 ft (rounded)
* DING, DING, DING: jbarrika was right!!!

For some reason some people weren't able to get this answer by using their calculators, this is how I recomend you to do it:
1- (.5)^10= .999....
ANS/ (1-(.5))= 1.998....
ANS (100)= 199.8
ANS+ 100= 299.8


Now, some new challenges...

2.) Find the sum of the first 10 terms of the following geometric series:
5+ 25+ 125...
Isa's Solution:
r= 25/5= 5
*r is the ratio
Sn= tn ((1-r^n)/(1-r))
S10= 5 ((1-5^10)/(1-5))= 12.207030

3.) If t1= 1, t2= -3, find S20
a) If the series is Geometric
b) if the series is Arithmetic

a.) r= -3/1 = -3
Sn= t1 ((1-r^n)/(1-r))
S20= 1 ((1- (-3)^20)/(1- (-3))
= -3, 486,784,399

b.) d= -4
tn= t1+(n-1)d
t20= 1+ (20-1) 2
= 39
S20= 20/2 (1+39)
= 10 (40)
= 400