Mathematical Proof Recurrence Relation

Question

How can i proof the below recurrence

$T_1 = 3,~~~~~~$ $T_n = T_{n-1} + 2$

is equal to this function $T_n = 2n + 1$?

When n= 1, T1= 3= 2(1)+ 1. Suppose that, for some k Tk= 2k+ 1. Then T(k+1)= (2k+1)+ 2= (2k+ 2)+ 1. — user247327, Jun 10 '17 at 15:51
I understand that the word recurrence needs a demonstration but it should be said also that it is an Arithmetic Progression and we could have used the classical formula for the general term. — Jean Marie, Jun 10 '17 at 15:58
See https://math.stackexchange.com/questions/2315846/find-closed-form-for-a-1-2-a-n-a-n-1n6/2315853#2315853 — lab bhattacharjee, Jun 10 '17 at 16:00

score 1 · Answer 1 · answered Jun 10 '17 at 15:53

1

Let $P_n=2n+1$

Starting from $T_n=T_{n-1}+1$, we have

$$T_n-P_n=T_n-2n-1=T_{n-1}+2-2n-1=T_{n-1}-2(n-1)-1=T_{n-1}-P_{n-1}$$

So, the difference between the two sequences remains constant. Therefore, we have

$$T_n-P_n=T_1-P_1$$

answered Jun 10 '17 at 15:53

CY Aries

score 0 · Answer 2 · answered Jun 10 '17 at 15:51

Let $S=\{n\in\Bbb N\setminus\{0\}\,:\, T_n\ne 2n+1\}$ and assume as a contradiction that $S\ne \emptyset$. Consider $m=\min S$.

Clearly, $m\ne 1$, because $T_1=3$.

However, if $m\ge 2$, then $T_{m-1}=2m-1$ and $T_m=T_{m-1}+2=2m-1+2=2m+1$. This cotradicts $m\in S$, thus proving $S=\emptyset$. QED.

2 Answers2