Question

Population growth/geometric series help. *question attached below

Answer 1

So can someone just explain to me how to do this question. All I want is a guide and then I'll tackle it :)

Answer 2

so, you would subtract 100 before you add a 20% increase. (Since 100 ppl. leave on Dec 31, and the year isn't yet over).

Answer 3

The entire value is \(\large\color{navy}{ 100\text{%} }\).
And \(\large\color{navy}{ +20\text{%} }\) means \(\large\color{navy}{ 120\text{%} }\).

So you are taking \(\large\color{navy}{ 120\text{%} }\) each time.
That means, you are multiplying times \(\large\color{navy}{ 1.2 }\) every time. \(\large\color{navdy}{ ( }\) i.e. \(\large\color{navy}{ r=1.2 }\) \(\large\color{navdy}{ ) }\)

Answer 4

can you (simply) find \(\large\color{navy}{ p_1 }\) \(\large\color{navdy}{ ( }\) i.e the population after the first year \(\large\color{navdy}{ ) }\) ?

Answer 5

Let me attempt it

Answer 6

1000(1.2)-100 = 1100

Answer 7

as I said, 100 people leave before the year is over. So your order of operations is off.

Answer 8

(1000-100)(1.2)?

Answer 9

you first subtract 100 \(\large\color{navdy}{ ( }\)i.e \(\large\color{navy}{ 100 }\) people leave on Dec 31\(\large\color{navdy}{ )}\)

Answer 10

yes, the last thing you said is right.

Answer 11

So p1=1080

Answer 12

yes.

Answer 13

now, the same way, find \(\large\color{navy}{ {\rm p}_2 }\) \(\large\color{navdy}{ ( }\) starting from \(\large\color{navy}{ {\rm p}_1 }\) \(\large\color{navdy}{ ) }\)

Answer 14

(1080-100)(1.2)=1176

Answer 15

\(\large\color{navy}{ \left({\rm p}_1 -100\right)\times1.2 }\)
\(\large\color{navy}{ \left(1080 -100\right)\times1.2 }\)

yes correct.

Answer 16

Now, you need part B.

Answer 17

what pattern do you see? \(\large\color{navy}{ (}\) Saying, what is the pattern we used to go:
1) from \(\large\color{navy}{ {\rm p}_0 }\) to \(\large\color{navy}{ {\rm p}_1 }\)
2) from \(\large\color{navy}{ {\rm p}_1 }\) to \(\large\color{navy}{ {\rm p}_2 }\) \(\large\color{navy}{ )}\)

Answer 18

Hint: `you are subtracting 100 and then multiplying times 1.2 `

Answer 19

It just wants the relation between \(\large\color{navy}{ {\rm p}_{n+1} }\) and \(\large\color{navy}{ {\rm p}_{n} }\)

Answer 20

From P0 to P1: (P0-100)(1.2)
From P1 to P2: (P1-100)(1.2)

Answer 21

yes, so....

Answer 22

So now: Pn+1= (Pn-100)(1.2) ?

Answer 23

yes, for part b, it is: \(\large\color{navy}{ {\rm p}_{n+1} =\left({\rm p}_{n}-100\right)\times1.2}\)

Answer 24

Now, part C, the most complicated one.

Answer 25

I am not very sure how to prove this, i might want to think a bit more...

Answer 26

No problem! I'll be thinking of something too

Answer 27

what do they want you to do, when they ask you to show it?
Do they want to plug in values into both patterns and evaluating the \(\large\color{navy}{ {\rm p}_{\rm something}}\) ?

Answer 28

I guess they want us to use steps to finally arrive at Pn, considering the question is worth 6 marks

Answer 29

I was incorrect most likely. http://www.wolframalpha.com/input/?i=500%281.2%29%5E1%2B500%3D%281000-100%29%281.2%29%5E1

Answer 30

this pattern in part C is wrong according to what I said (for n=1 at least), so I guess you subtract 100 after multiplying

Answer 31

\(\large\color{royalblue}{ {\rm p}_1=(1000\times1.2) -100 =1100 }\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\large\color{royalblue}{ {\rm p}_2=(1100\times1.2) -100 =1220 }\)

Answer 32

\(\large\color{royalblue}{ {\rm p}_{n+1} =\left({\rm p}_{n} \times 1.2\right) -100 }\)

Answer 33

But if we use the formula given in C, it doesn't add up :S

Answer 34

what do you mean?

Answer 35

Ahh, sorry. I made a mistake. You are correct.

Answer 36

you are made a mistake saying that I was correct.... (lol)
yes, you are subtracting 100 after multiplying times 1.2

Answer 37

Or else, then the terms get smaller and smaller.

Answer 38

\(\large\color{royalblue}{ {\rm p}_{n+1} =\left({\rm p}_{n} \times 1.2\right) -100 }\)

Answer 39

Now, we need to show that: \(\large\color{royalblue}{ {\rm p}_{n+1} =500\left(1.2 \right)^n+500 }\)
it is still hard, but not as hard.

Answer 40

Hmm

Answer 41

I can't really think of anything besides observing how each pattern goes.

Answer 42

What I mean is, just to plug in vlaues into both patterns.

Answer 43

I will try one more time...

Answer 44

I got some latex I am on my way

Answer 45

No problemo :)

Answer 46

\(\large\color{royalblue}{ \displaystyle {\rm p}_1}\)

\(\large\color{royalblue}{ \displaystyle (1000)(1.2)-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)+(500)(1.2)-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)+600-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)+500}\)


\(\large\color{royalblue}{ \displaystyle {\rm p}_2}\)

\(\large\color{royalblue}{ \displaystyle ((500)(1.2)+500)(1.2)-100}\)
\(\large\color{royalblue}{ \displaystyle ((500)(1.2)^2+600)-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)^2+600-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)^2+500}\)


\(\large\color{royalblue}{ \displaystyle {\rm p}_3}\)

\(\large\color{royalblue}{ \displaystyle ((500)(1.2)^2+500)(1.2)-100}\)
\(\large\color{royalblue}{ \displaystyle ((500)(1.2)^3+600)-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)^3+600-100}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)^3+500}\)
\(\large\color{royalblue}{ \displaystyle (500)(1.2)+500}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\large\color{royalblue}{ \displaystyle \sum_{ n=1 }^{ \infty }\left[ ~~500(1.2^n)+500\color{white}{{\huge|}} \right] }\)

Answer 47

so you can see the pattern here is that each time the power to 1.2 gets added , and the +500 remains same.

Answer 48

and the
\(\large\color{royalblue}{ \displaystyle \sum_{ n=1 }^{ \infty }\left[ ~~500(1.2^n)+500\color{white}{{\huge|}} \right] }\)

is what the pattern sums to.

Answer 49

Sorry it took too long for me to figure out....
but once you know that:
1) \(\large\color{slate}{a(b+c)~~~~~\Longrightarrow~~~~~ab+ac }\)

2) \(\large\color{slate}{500\cdot1.2=600 }\)

3) \(\large\color{slate}{(1.2)^{anything}~~~\cdot(1.2)=(1.2)^{anything+1} }\)

it shouldn't be hard.

Answer 50

go over my comments, carefully, tnx and apologize for my brainless ness. I am not good at math.

Answer 51

there is a typo when I said \(\large\color{slate}{p_3 }\) in a last line

Answer 52

it should have 1.2^3

Answer 53

are you clear with what I said for part C?

Answer 54

Thank you very much! I'm looking them over now and I'm definitely understanding. And you're so smart to have come up with this! Many thanks again!

Answer 55

Yes, I am very clear :)

Answer 56

It is just a sigma.. the pattern is like this... don't know why I struggled so much with this prob.
Yw