Is there a simple way to find the recursive equation for a quadratic sequence?
$\endgroup$ 22 Answers
$\begingroup$If by a quadratic sequence you mean the values of a quadratic polynomial at integers, then consider the sequence of differences. The sequence of second differences is constant and so the sequence of first differences is an arithmetic progression, for which there is a simple formula. A recursive equation for the original quadratic sequence is then easy.
More precisely, if the quadratic sequence is given by $q(n)$, where $q$ is a quadratic polynomial, then $d(n)=q(n+1)-q(n)$ is the arithmetic progression given by $d(n)=an+b$, where $a$ is the second difference and $b=d(0)$. So, $q(n+1)=q(n)+d(n)=q(n)+an+b$, where $a=d(1)-d(0)=q(2)-2q(1)+q(0)$ and $b=d(0)=q(1)-q(0)$.
Your example is given by $q(n+1)=q(n)+2n+3$, $q(0)=4$.
$\endgroup$ $\begingroup$one more easy formula would be t1+(n-1)(nx/2+t2-t1-x) where t1= first term t2= second term x= second difference. In the example you gave t1=4, t2=7, x=2. By applying the formula we get: 4+(n-1)(2n/2+7-4-2) 4+(n-1)(n+1) 4+n^2-1 = n^2+3 This would be the nth term. In case you don't know second difference it the difference between the difference in the series. Like the difference of 4 and 7 is 3 and the difference of 7 and 12 is 5, so the second difference is the difference between 3 and 5 which is 2. Hope this helps
$\endgroup$
