Combinatorics back-substitution Question

[Askhwhelp]

1) an = an-1 + 2n with a0 = 2
The question is Use back-substitution to find a closed formula for the recurrence relations.
I have an-4 + 2(n-3)+2(n-2)+2(n-1)+2n
Then I have a1 + ... +2(n-5)+2(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
First of all, is my way right? If so, could you please help me find the recurrence relations?
If no, could you tell I have done wrong and point me to the right direction?

2) Find and solve a recurrence relation for pn the value of a stock market indicator that obeys the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's
change. Let p0 = 1; p1 = 4.
I have a trouble to translate " the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's" into recurrence relation. Can someone please help?

 

[LCKurtz]

1) an = an-1 + 2n with a0 = 2
The question is Use back-substitution to find a closed formula for the recurrence relations.
I have an-4 + 2(n-3)+2(n-2)+2(n-1)+2n
Then I have a1 + ... +2(n-5)+2(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
First of all, is my way right? If so, could you please help me find the recurrence relations?
If no, could you tell I have done wrong and point me to the right direction?

I presume you mean $a_n = a_{n-1} +2n$ with $a_0=2$. I would write out the first few, taking care not to simplify, which might hide any patterns:

$a_1 = 2 + 2\cdot 1$
$a_2 = 2 + 2\cdot 1 + 2\cdot 2$
$a_3 = 2 + 2\cdot 1 + 2\cdot 2 + 2\cdot 3$

Do you see any patterns or useful expressions you can use to get a nice closed formula for $a_n$?

 

[Askhwhelp]

The answer should be 2+n+n^2?

Thanks

 

[ArcanaNoir]

2) Find and solve a recurrence relation for pn the value of a stock market indicator that obeys the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's
change. Let p0 = 1; p1 = 4.
I have a trouble to translate " the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's" into recurrence relation. Can someone please help?

Sounds to me like: $(p_2-p_1)=2(p_1-p_0)$ so, $p_n-p_{n-1}=2(p_{n-1}-p_{n-2})$.

 

[LCKurtz]

The answer should be 2+n+n^2?

Thanks

Can't you check it for yourself?