recurrence relation

**papa_chango123** · December 21st 2006, 06:31 PM

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

**papa_chango123** · December 21st 2006, 06:32 PM

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

**ThePerfectHacker** · December 21st 2006, 07:10 PM

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

**Soroban** · December 22nd 2006, 12:05 PM

Hello, papa_chango123!

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

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

Am I reading it correctly?

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

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

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

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

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

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

**papa_chango123** · December 22nd 2006, 01:20 PM

Yes thanks that is the answer I got.

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