Answer
The second order differences are equal to $3$.
Work Step by Step
We are given:
$$f(n)=\dfrac{1}{2}n(3n-1).$$
We calculate the general term of the first order differences:
$$\begin{align*}
f(n+1)&=\dfrac{1}{2}(n+1)(3(n+1)-1)\\
&=\dfrac{1}{2}(n+1)(3n+2).\\
f(n)&=\dfrac{1}{2}n(3n-1).\\
f(n+1)-f(n)&=\dfrac{1}{2}(n+1)(3n+2)-\dfrac{1}{2}n(3n-1)\\
&=\dfrac{1}{2}(3n^2+5n+2-3n^2+n)\\
&=\dfrac{1}{2}(6n+2)\\
&=3n+1.
\end{align*}$$
Calculate the general term of the second order differences:
$$\begin{align*}
3(n+1)+1-(3n+1)=3n+3+1-3n-1=3.
\end{align*}$$
We got that the second order differences are constant.
