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=\frac{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`.