# recurrence relation

• Dec 21st 2006, 06:31 PM
papa_chango123
recurrence relation
Can anyone tell me if this is correct for this relation:

6.1 In the following sequences determine s5 if s0, s1, ... sn, ... is a sequence satisfying the given recurrence relation and initial condition.
b. sn = -sn-1 - n2 for n >= 1, s0 = 2
• Dec 21st 2006, 06:32 PM
papa_chango123
Sorry forgot to put in the answers. Here is the answers I got:

s0 = 2
s1 = -1
s2 = 5
s3 = 4
s4 = 12
s5 = 13
• Dec 21st 2006, 07:10 PM
ThePerfectHacker
Quote:

Originally Posted by papa_chango123
Sorry forgot to put in the answers. Here is the answers I got:

s0 = 2
s1 = -1
s2 = 5
s3 = 4
s4 = 12
s5 = 13

No,
s1=-s0-1^2=-2-1=-3
• Dec 22nd 2006, 12:05 PM
Soroban
Hello, papa_chango123!

What is $\displaystyle n2$?
Is that your way of writing $\displaystyle n^2$ ?

Quote:

6.1) In the following sequences determine $\displaystyle s_5$
if $\displaystyle s_n$ is a sequence satisfying the given recurrence relation and initial condition.
$\displaystyle b)\;s_n \:= \:-s_{n-1}- n^2$ . for $\displaystyle n \geq 1,\;s_o = 2$

It says: .$\displaystyle \underbrace{s_n}_{\text{the nth term}} \;\underbrace{=}_{\text{is}}\;\underbrace{-s_{n-1}}_{\text{neg.of preceding term}} - \underbrace{n^2}_{\text{minus n-squared}}$

So $\displaystyle s_1$ is the negative of $\displaystyle s_o$, minus $\displaystyle 1^2.$
. . $\displaystyle s_1\:=\:-2 - 1^2\:=\:-3$

And $\displaystyle s_2$ is the negative of $\displaystyle s_1$, minus $\displaystyle 2^2.$
. . $\displaystyle s_2\:=\:-(-3) - 2^2 \:=\:-1$

Then $\displaystyle s_3$ is the negative of $\displaystyle s_2$, minus $\displaystyle 3^2.$
. . $\displaystyle s_3\:=\:-(-1) - 3^2 \:=\:-8$

Hence $\displaystyle s_4$ is the negative of $\displaystyle s_3$. minus $\displaystyle 4^2.$
. . $\displaystyle s_4\:=\:-(-8) - 4^2\:=\:-8$

Therefore: $\displaystyle s_5$ is the negative of $\displaystyle s_4$, minus $\displaystyle 5^2.$
. . $\displaystyle s_5\:=\:-(-8) - 5^2\:=\:\boxed{-17}$

• Dec 22nd 2006, 01:20 PM
papa_chango123
Yes thanks that is the answer I got.