1. ## Proving

Each year in June approximately 10% of the trees die out and in December, the workers plant 100 new trees. At the end of December, the workers plant 100 new trees. At the end of December 2000 there are 1200 trees in the plantation.

By considering 2001 as the first year. Show that the number of living trees at the end of December in the nth year is given by $(\frac{9}{10})^n(1200)+1000(1-0.9^n)$

No. of living trees

$First Year=0.9(1200)+100$

$Second Year=0.9(0.9(1200)+100)+100=0.9^2(1200)+0.9(100)+1 00$

No idea where the 1000 came about
$Third Year=0.9(0.9(0.9(1200)+100)+100)+100=0.9^3(1200)+0 .9^2(100)+100$

For the $n^{th} year=0.9^n(1200)+0.9^{n-1}(100)+100$

2. use mathematical induction theory. When doing that use (1-0.9) as 0.1 it will make proving for n=p+1 easier.

Spoiler:

assume true for n=p
then,
$S_p=(0.9)^p(1200)+1000(1-0.9^p)$

$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p)+100-0.1[(0.9)^p(1200)+1000(1-0.9^p)]$
$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -(1-0.9)(0.9^p(1200))-1000(1-0.9)(1-0.9^p)$
$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -0.9^p(1200)+0.9^{p+1}(1200)-1000((1-0.9^p)-0.9+0.9^{p+1})$
$S_{p+1}=1000(1-0.9^p)+100 +0.9^{p+1}(1200)-1000(1-0.9^p)+900-1000(0.9^{p+1})$
$S_{p+1}=100 +0.9^{p+1}(1200)+900-1000(0.9^{p+1})$
$S_{p+1}=0.9^{p+1}(1200)+1000-1000(0.9^{p+1})$
$S_{p+1}=0.9^{p+1}(1200)+1000(1-0.9^{p+1})$

so the statement is true for n=p+1

use mathematical induction theory. When doing that use (1-0.9) as 0.1 it will make proving for n=p+1 easier.

Spoiler:

assume true for n=p
then,
$S_p=(0.9)^p(1200)+1000(1-0.9^p)$

$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p)+100-0.1[(0.9)^p(1200)+1000(1-0.9^p)]$
$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -(1-0.9)(0.9^p(1200))-1000(1-0.9)(1-0.9^p)$
$S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -0.9^p(1200)+0.9^{p+1}(1200)-1000((1-0.9^p)-0.9+0.9^{p+1})$
$S_{p+1}=1000(1-0.9^p)+100 +0.9^{p+1}(1200)-1000(1-0.9^p)+900-1000(0.9^{p+1})$
$S_{p+1}=100 +0.9^{p+1}(1200)+900-1000(0.9^{p+1})$
$S_{p+1}=0.9^{p+1}(1200)+1000-1000(0.9^{p+1})$
$S_{p+1}=0.9^{p+1}(1200)+1000(1-0.9^{p+1})$

so the statement is true for n=p+1
But i'm supposed to produce the equation and not show that the equation shown is true.
However, you started off with the equation given, could you have misunderstood the question?

4. I guess it is the nature of the mathematical induction theory. You can take a statement and show it is true (I think you know mathematical induction theory)

I guess it is the nature of the mathematical induction theory. You can take a statement and show it is true (I think you know mathematical induction theory)
Sorry, but i dont think i can use mathematical induction theory (because i havent learnt it before so it is not in my syllabus)

6. $(0.9)^n(1200)+1000(1-0.9^n)$

$\\First\ Year = 0.9(1200) + 100\\Second\ Year = 0.9(0.9(1200)+100)+100\\Third\ Year = 0.9(0.9(0.9(1200) + 100)+100)+100$

$\\First\ Year = 0.9(1200) + 100\\Second\ Year=0.9^2(1200)+0.9(100)+100\\Third\ Year=0.9^3(1200)+0.9^2(100) + 0.9(100) + 100\\Nth\ Year= 0.9^n(1200)+(0.9)^{0}(100)+0.9^{1}(100)+...+0.9^{n-2}(100)+0.9^{n-1}(100)$

So the $(0.9)^n(1200)$ part should be obvious.

Let's show that:
$0.9^{0}(100)+0.9^{1}(100)+...+0.9^{n-2}(100)+0.9^{n-1}(100) = 1000(1-0.9^n)$

Take out a thousand:
$1000(\ \ 0.9^{0}(1/10)+0.9^{1}(1/10)+...0.9^{n-2}(1/10)+0.9^{n-1}(1/10)\ \ )$
Take out a tenth:
$1000(\ \ (1/10)(1+0.9^1+0.9^2+...+0.9^{n-2}+0.9^{n-1})\ \)$

Geometric Series Found!

$1000 \left((1/10)\left(\frac{1-0.9^n}{1-0.9}\right)\right)=1000(1-0.9^n)$

Good luck!