Question

Find the recursive quadratic formula of the sequence: 1, 3, 7, 13, 21

Answer 1

The sequence

1, 3, 7, 13, 21, ...

has first-order differences

2, 4, 6, 8, ...

Let $a_n$ denote the original sequence, and $b_n$ the sequence of first-order differences. It's quite clear that

$b_n=2n$

for $n\ge1$. By definition of first-order differences, we have

$b_n=a_{n+1}-a_n$

for $n\ge1$, or

$a_{n+1}=a_n+2n$

By substitution, we have

$a_n=a_{n-1}+2(n-1)$

$\implies a_{n+1}=(a_{n-1}+2(n-1))+2n$

$\implies a_{n+1}=a_{n-1}+2(n+(n-1))$

$a_{n-1}=a_{n-2}+2(n-2)$

$\implies a_{n+1}=(a_{n-2}+2(n-2))+2(n+(n-1))$

$\implies a_{n+1}=a_{n-2}+2(n+(n-1)+(n-2))$

and so on, down to

$a_{n+1}=a_1+2(n+(n-1)+\cdots+2+1)$

You should know that

$1+2+\cdots+(n-1)+n=\dfrac{n(n+1)}2$

and we're given $a_1=1$, so

$a_{n+1}=1+n(n+1)=n^2+n+1$

or

$a_n=(n-1)^2+(n-1)+1\implies\boxed{a_n=n^2-n+1}$

Alternatively, since we already know the sequence is supposed to be quadratic, we can look for coefficients $a,b,c$ such that

$a_n=an^2+bn+c$

We have

$a_1=a+b+c=1$

$a_2=4a+2b+c=3$

$a_3=9a+3b+c=7$

and we can solve this system for the 3 unknowns to find $a=1,b=-1,c=1$.