# Math Help - Proof: recursive and closed formulas

1. ## Proof: recursive and closed formulas

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2. 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 $a_n =2^{n-1}-3$.

3. 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?

4. sort of..i'm in the process of learning it. how does it work?

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

6. ..

7. 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 $a_{n+1}=2a_n+3=2\left(2^{n-1}+3\right)-3=2^n+6-3=2^n+3$

8. thanks that makes sense. i also just found a really stupid error in my work.
got it now!