Jefferson's Scribe post

the next scribe is JHO-ANN.

a) what is the sum of the integers from 1 to 5000?

T₁ = 1
D = 1
N = 5000

‘the initial term is 1
‘the to from getting from 1 to 5000 is 1. on account that each term increases by one. And that it doesn’t say what the difference really is so its 1
‘n is the rank of the rank of the term we want to find. The rank is the sum in this case and we want to find what the sum is when the sequence reaches 5000.

S_n = n/2[2(t1) + (n – 1)d]
S₅₀₀₀ = (5000/2)[2(1) + (5000 – 1)1]
S₅₀₀₀ = 2500[2 + 4999]
S₅₀₀₀ = 2500 = [5001]
S₅₀₀₀ = 12 502 500

b) what is the sum of all multiple of 7 between 1 & 5000?

T₁ = 7
D = 7
N = 714

Since it says multiples of 7,the initial term is 7 because you start with 7 when it comes to “multiples of 7”. The difference is 7 because it increases by 7 each time.
Now we want to find the sum of sequence when it reaches to 5000. but we don’t really want to find 5000. since we’re looking for the sum of the multiples 7 between 7 & 5000 then there really aren’t 5000 terms if we’re increasing by 7 each time. To find what the rank is, we need to 5000/7. and you get 714. whatever. Don’t need to worry about the decimal answers if I were you.

S_n = n/2[2(t1) + (n – 1)d]
S₇₁₄ = (714/2)[2(7) + (714 – 1)7]
S₇₁₄ = 357[(14 + 7130)]
S₇₁₄ = 1 784 785

c) What is the sum of all integers from 1 to 5000 inclusive that are not multiple of 7?

12 502 500 – 1 786 785
= 10 715 715

This is a no brainer but I’ll explain it to you anyways. To get the answer you take the sum of all integers from 1 to 5000 and subtracted by the sum of the multiples of 7.

A super ball is dropped from a height of 200 cm. It rebounds to ¾ of the distance it fell each time it hits the ground. What is the total vertical distance traveled by the ball when it hits the ground for the fourth time?

We know that each time it hits the ground,the ball comes back ¾ of the last distance it traveled. So it traveled down, up, down, up, down, up, and down. So in order to calculate the total distance it traveled. We need to brake this into 2 solutions first. We need to solve the “down” distance and then the “up” distance and then add the together.

Down:

So what are the sequences?

Well since the first term is 200 and it decreases every time by a ¼( bounces back to ¾ of the last term) and that it hits the ground for times…..

200 + 150 + 112.5 + 84.375

(but we really didn’t need to know that only the first term)

*DID YOU KNOW: that all the things we need to know to solve the question was in the question itself? Go check it out ; - )

T₁ = 200
R = ¾
N = 4

S₄ = 200(1 – (3/4)⁴)/(1 – ¾)
S₄ = 546.875

(thanks to the help of our righteous hero- the CALCULATOR!!)

The 3 term for the “down” distance is:
‘think everything opposite of what I said from the other part

150 + 112.5 + 84.375

T₁ = 150
R = ¾
N = 3

S₃ = 150( 1 – (3/4)³)/ (1 - ¾)
S₃ = 346.875

Now we add the two answer to get …….

546.875 + 346.875 = 893.75 (anybody else noticed that the is a difference of 200 for the 2 answers. Me thinks there is a shortcut for this problem….)

……………TAAADAAAAA!!

I've voice recorded mr k's lecture today. It covered about half the class. Now i have listened to it a bit and you have to crank up the volume super high to listen on account that the mic was far from mr k. I just wanted to test out if it worked or not. Feel free to add some comments about it. Apologies to the loud scratching sounds on account i was moving the mp3 play on the desk, i assumed that it picked up the sound.


test voice lecture