Scribe Post (May 11)

As usual, Mr. A. started the class with a POD

POD 5/11

a) Trace your direct ancestors back ten generations. How many people does that include?(ignore any inbreeding)

b) How many generations must you go back to find 1 million ancestors?

Jaime was the first one to finish the problem, so he was called up to the board.

First he showed us how the sequence works:
2,4,8,16,32......

Then he used the geometric sequence formula to solve the problem:

T1=2, r=2
Sn=t1((1-(r^n))/(1-r))
S10=2((1-(2^10))/(1-2))
S10=2,046 ancestors

Then Juliana was called up to the board to do part b. She used the geometric SEQUENCE formula to find the "answer":

tn=t1(r^(n-1))
1,000,000/2=2(2^(n-1))/2
log2500,000=log2(2^(n-1))
log2500,000=n-1

Juliana didn't know how to solve for n, so Mr. A. had to show us:

y=log2500,000
ln2(y)=ln500,000/ln2
y=18.93156857

This answer is slightly wrong because she used the geometric sequence formula to solve a geometric series problem. Isabella showed us how to do this problem correctly by using the geometric series formula:

Sn=t1((1-(r^n))/(1-r))
1,000,000/2=2((1-(2^n))/(1-2))/2
500,000(-1)=(1-(2^n))/-1(-1)
-500,000(-1)=(1-(2^n))(-1)
500,000+1=-1+(2^n)+1
500,001=2^n
log2500,001=log2(2^n)
n=log2500,001
n=ln500,001/ln2
n=18.93157145

Then Mr. A. put a problem on the board and showed us how to do it:

t1=1, t2=-3, find S20
a) In a geometric series.
b) In an arithmetic series.

a)
Sn=t1((1-(r^n))/(1-r))
S20=1((1-(-3^20))/(1-(-3))
S20=-871,695,100.5

b)
d=t2-t1
d=-4
tn=t1+((n-1)(d))
t20=1+((19)(-4))
t20=-75
Here Mr. A. used Gauss´s technique to solve the problem
S20=20/2(1-75)
S20=-740

Then Mr. A. gave us another problem to solve:

A grain of salt is placed on the corner square of a large chess board. Two grains are placed on the next square, four grains on the next, eight on the next and so on.

a) How many grains will the board contain after half the squares have been covered?
b.) When all squares are covered?

Francisco went up to the board to to the problem. He showed us how the sequence works and he used the geometric series formula to solve the problem:

1,2,4,8,16......
Sn=t1((1-(r^n))/(1-r))
S32=1((1-(2^32))/(1-2))
S32=4,294,997,295

S64=1((1-(2^64))/(1-2))
S64=1.844674 x 10^19

Then Mr. A. solved for t16 and t32 on the same problem. He used the geometric sequence formula to solve it:

tn=t1(r^(n-1))
t16=1(2^15)
t16=32,768 grains

t32=1(2^31)
t32=2,147,483,684

Towards the end of the class Mr A gave us an Assigment:

Create both an arithmetic and geometric series word problem.

• Post your problems without solutions on the blog by midnight tonight.
  

THE CLASS ENDS!!!