Problem
NT-B2-M12-P005 Second Difference
#5
★★☆☆☆ Level 2 of 5
For \(u_n=3n^2+2n+1\), prove that \(u_{n+1}-2u_n+u_{n-1}\) is constant, and find this constant value.
Substitute \(n+1\), \(n\), \(n-1\), or recall that the second difference of a quadratic polynomial is constant.
Compute: \(u_{n+1}=3(n+1)^2+2(n+1)+1=3n^2+8n+6\), and \(u_{n-1}=3(n-1)^2+2(n-1)+1=3n^2-4n+2\). Then \(u_{n+1}-2u_n+u_{n-1}=6\). This is \(2\) times the leading coefficient \(3\).
A small entry point into finite differences.