[Math] The sum of the first $n$ terms of an arithmetic sequence is given by $3n+2n^2$. find the common difference of the sequence

Question

The sum of the first $n$ terms of an arithmetic sequence is given by $3n+2n^2$. find the common difference of the sequence.

So the answer is 4, but I've got 2 – my workings are shown – is my method wrong or have i just made a mistake.

$$d=U_n-U_{n-1}$$
$(1)$ $$U_1+U_2+…+U_{n-1}+U_n=3n+2n^2$$
$(2)$ $$U_1+U_2+…+U_{n-1}=3(n-1)+2(n-1)^2$$
$(1)-(2)$ = $U_n=2n-2$

This would give me a sequence of 0,2,4 which has a common difference of 2 not 4 Any help would be appreciated.

Answer 1

Getting to your calculation $2n^2 + 3n - 2(n-1)^2 - 3(n-1) = 2n^2 + 3n -2n^2 + 2*2n - 2 - 3n +3$ which gives $U_n=4n+1$. So it looks like you made an arithmetic error.