Given the following definition, compute Q(5).
Q(n) =
0 if n = 0
1 if n = 1
3 if n = 2
Q(n − 1) + Q(n − 2) + Q(n − 3) if n 2
Q(5) =

Question

Given the following definition, compute Q(5).
Q(n) = 
0	if n = 0
1	if n = 1
3	if n = 2
Q(n − 1) + Q(n − 2) + Q(n − 3)  	if n > 2
Q(5) =

anonymous · Answer

i got it..i asked this one before lol

loser66 · Answer

ok,

anonymous · Answer

actually one of the numbers is diff. its a 3 instead of 2..

anonymous · Answer

is the answer still 11?

loser66 · Answer

oh, got what you mean, you mean $$Q_n = Q_{n-1}+ Q_{n-2}+Q_{n-3}$$ if n >2

loser66 · Answer

so, $$Q_4 = Q_3+Q_2+Q_1= 3$$
 and $$Q_5 = Q_4+Q_3+Q_2= 7$$
 but it is really elementary step, love,
It is not our level. In discrete math, you have to find out the general form first, cannot take step lke that, right?

anonymous · Answer

since the value of each know q is its index ... that makes it simpler even q5 = ((q2+q1+q0)+q2+q1)+(q2+q1+q0)+q2 q5 = 2+1+0+2+1+2+1+0+2
??

loser66 · Answer

oops,

anonymous · Answer

oh got it its 15! :) i just replaced the 2s with 3s

anonymous · Answer

thanks!

loser66 · Answer

no no, I don't get what you mean, how?

anonymous · Answer

q5 = 3+1+0+3+1+3+1+0+3

loser66 · Answer

@e.mccormick

loser66 · Answer

why did she get that formula,friend?

e.mccormick · Answer

She is showing the totality of it.  Because $Q_5=Q_4+Q_3+Q_2$ but $Q_4=Q_3+Q_2+Q_1$ and $Q_3=Q_2+Q_1+Q_0$ therefore $Q_5=(Q_3+Q_2+Q_1)+(Q_2+Q_1+Q_0)+Q_2$

loser66 · Answer

Yeah, got it, thank you.

e.mccormick · Answer

=)  Have fun!

loser66 · Answer

ok XD