Scribe 6/5

Mr A started our class as usual, with a POD.

POD 6/5:
Find, in radians, all 1st revolution angles, where cot = - √3.
(NO CALCULATORS OR SNOWMAN SHEET)

Mr A´started us off by saying how he would have solved this problem. He told us how he would ignore the √3 and focus on the negative sign. The tan / cot angles are negative in both the second and fourth quadrants.

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*****REMINDER: In the final exam 1/3 of the quiz will be without a calculator
2/3 of the quiz will be with a calculator

Going back to what we were doing yesterday...
We left a problem without solution:

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Now that we have solved this problem, we are ready to prove the first properties we were given yesterday, by solving an exercise we have done before:

3+24+1.92+1.536+... FIND the sum of the terms 5 - 18

Answer: S5-18≈ 5.87


Find:

18
∑3*0.8^k-1
k:5

Explicit formula T1*r^k-1
(Geometric Sequence)

1st property (for further info. check Daniel's scribe)
18 18
∑3*0.8^k-1 = 3∑0.8^k-1
k:5 k:5



By using the 2nd property we will brake the equation into 2 geometric series eliminating the first four terms giving us as a result the sum of the terms 5 - 18

18 18 4
3∑0.8^k-1 = 3[(∑0.8^k-1) -(∑0.8^k-1)]
k:5 k:1 k:1

Use the Geometric series formula to evaluate:
Geometric Series formula Sn: T1(1-r^n/1-r)
18 4
3[(∑0.8^k-1) -(∑0.8^k-1)]
k:1 k:1


T1= 1 (FACTOR THE 3 OUT)
r= .8

3[1*(1-.8^18/.2)-1*(1-.8^4/.2)] = 15 (-.8^18+ .8^4) ≈ 5.87

5.87= 5.87

Mr A proved us that solving this series problem with sigma notation does work, and told us that it would be more useful to solve more complex exercises.

Solving another problem:
Find:

20 20 9
∑2n = 2[∑n-∑n]
n:10 n:1n:1

2[(20/2(21))-(9/2(10))]
2(10(21)-(45))
2(210-45)
=330

TIP: List first terms if you doubt which type of series you are being asked


Practicing with another problem...

30 30 30
∑3p-∑9p = ∑-6p
p:1 p:1 p:1

30
=-6∑p
p:1

=-6(15)(31)

=-2790


MORE PROPERTIES

3.The sum of the 1st n positive integers.

n
∑+=n(n+1)/2
i:1


4. The sum of the squares of the 1st n + integers. (For further info. look at Jaime´s post)
n
∑i^2=n(n+1)(2n+1)/6
i:1


Proving property #4
n
∑i^2=n(n+1)(2n+1)/6 [expand and add]
i:1

4(5)(9)/6 = 180/6 = 30

With property #4 we can finish yesterday´s problem:

find

12 12
∑5k-k^2 - 390 - ∑k^2
k:1 k:1

n(2n-1)(n+1)/6

= 390- [12(25)13/6]
=390-650
=-260

390(12*13*25)/6
= -260



**** LAST REMEDIAL AT 6:15 ****