Find an explicit formula for the sequence defined recursively by $$a _{0}=7$$
$$a _{1}=10$$ $$a _{k}=2a _{k-1}-a _{k-2}$$

Question

mr.math · Answer

$$a(n)=3n+7$$.

anonymous · Answer

Your sort of guess for these right

mr.math · Answer

$$a_0=7$$
$$a_1=3+7$$
$$a_2=2(10)-7=20-7=2(3)+7.$$

mr.math · Answer

Find several $a_n$'s and try to see the pattern, Define a formula based on that pattern, and then prove it's true for all n in your domain.

anonymous · Answer

Can we prove with induction

mr.math · Answer

I think so.

anonymous · Answer

hmm

anonymous · Answer

so up to a_4

anonymous · Answer

my sequence from a_0 is 7,10,20,30,40

mr.math · Answer

How?

anonymous · Answer

oh wait

anonymous · Answer

totally forgot to subtract 7 in a_2

mr.math · Answer

$$a_0=7$$
$$a_1=10$$
$$a_2=20-7=13$$
$$a_3=26-10=16$$
$$a_4=32-13=19$$ ....

anonymous · Answer

ah ok

mr.math · Answer

You can see the pattern, right?

anonymous · Answer

not sure if i would have been able to spot it as fast as you did, but i see it now yea

mr.math · Answer

I think you can, maybe a couple of minutes slower :P At least you can spot it's a linear relation, and this is the easiest type.

anonymous · Answer

yea, nonlinear are harder

across · Answer

$$f(n)=3n+7,$$$$n\in\mathbb{N},$$$$\mathbb{N}=\left\{0,1,2,...\right\}.$$
Proof:
For n = 0,$$f(0)=3(0)+7=7$$which is true. Then assume it is true for n = k and prove that it is true for k+1.$$f(k+1)=f(k)+3,$$$$f(k+1)=3(k+1)+7=3k+3+7=3k+10=(3k+7)+3=f(k)+3.\blacklozenge$$yup