# Proof: recursive and closed formulas

• Feb 28th 2010, 01:14 PM
mathh18
Proof: recursive and closed formulas
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• Feb 28th 2010, 02:35 PM
Plato
Quote:

Originally Posted by mathh18
i'm trying to prove that recursive formula 2[a(sub)k-1] -3 and closed formula [a(sub)n]= 2^(n-1) +3 are equal.
i have [a(sub)m] = 2[a(sub)m-1]-3 =
2(2^(m-3)+3) -3 =
2^(m-2) +3
that is close, but slightly off than what i'm trying to prove..what should i do to make it 2^(m-1) +3 ?

sorry for the typing, it's confusing on the computer

You can learn to use LaTeX.
$$a_n =2^{n-1}-3$$ gives $a_n =2^{n-1}-3$.
• Feb 28th 2010, 03:20 PM
Drexel28
Quote:

Originally Posted by mathh18
i'm trying to prove that recursive formula 2[a(sub)k-1] -3 and closed formula [a(sub)n]= 2^(n-1) +3 are equal.
i have
[a(sub)m] = 2[a(sub)m-1]-3 =
2(2^(m-3)+3) -3 =
2^(m-2) +3
that is close, but slightly off than what i'm trying to prove..what should i do to make it 2^(m-1) +3 ?

sorry for the typing, it's confusing on the computer

I think this says prove that $a_n=2a_{n-1}+3\implies a_n=2^{n-1}+3$. Know induction?
• Feb 28th 2010, 03:38 PM
mathh18
sort of..i'm in the process of learning it. how does it work?
• Feb 28th 2010, 03:40 PM
Drexel28
Quote:

Originally Posted by mathh18
sort of..i'm in the process of learning it. how does it work?

• Feb 28th 2010, 03:54 PM
mathh18
..
• Feb 28th 2010, 05:24 PM
Drexel28
Quote:

Originally Posted by mathh18
haha yeah! that's what i'm doing. i made a chart for n=1 and n=2 n=m-1 to show its true for all of those values. then i tried proving it for m and got stuck

Much simpler than that. Note that if we assume that $a_{n}=2^{n-1}+3$ then $a_{n+1}=2a_n+3=2\left(2^{n-1}+3\right)-3=2^n+6-3=2^n+3$
• Feb 28th 2010, 05:49 PM
mathh18
thanks that makes sense. i also just found a really stupid error in my work. (Headbang)
got it now!