Show that the sequence defined by the formula $a_n=n+3$, satisfies the recurrence $a_n=2a_{n-1}-a_{n-2}$ for all $n\ge2$.

Question

Show that the sequence defined by the formula $a_n=n+3$, satisfies the recurrence $a_n=2a_{n-1}-a_{n-2}$ for all $n\ge2$.

I do understand that recurrence relation is recursion. I understand the idea, but I would just like to know how to prove this. I would really appreciate if you show me how to do this problem and give specific details. Thanks

Answer 1

Since the sequence is defined by $\forall n\ge 2,\ a_n=n+3,$ it would be a good idea to calculate the first few numbers in the sequence. So, then we have $$a_2=2+3=5,\quad a_3=3+3=6,\quad a_4=4+3=7,\quad a_5=5+3=8$$ As you can see, each $a_n$ is just one more than $a_{n-1}$. Then we have $$a_n=a_{n-1}+1\qquad(1)$$ Notice the difference between $a_{n-1}$ and $a_{n-2}$ is also $1$. So we have $$1=a_{n-1}-a_{n-2}\qquad (2)$$ Now, plug $(2)$ into $(1)$ and we arrive at $$a_n=a_{n-1}+a_{n-1}-a_{n-2}=2a_{n-1}-a_{n-2}$$

Answer 2

Just show $$2a_{n-1}-a_{n-2}=2(n-1+3)-(n-2+3)=n+3=a_n,\ n\ge 2$$