TITLE: Question about arithmetic progression. stuck on one of the answers
QUESTION [5 upvotes]: So i was doing this question on my workbook. I did the first question and i was correct. but on the second question, same logic, same solving way, but i was wrong.
here is question 1, finding a closed form for
$$a_1=3\qquad a_{n+1}=a_n+2,n\ge2$$
Thus I have
$$a_{n+1}=a_n+2,d=2$$
$$a_n=a_{n-1}+2$$
$$a_n=a_1+(n-1)*2$$
$$a_n=a_1+2n-2$$
$$a_1=3$$
$$a_n=3+2n-2$$
$$a_n=2n+1,n\ge1$$
the last line was the answer, and i did it right.but the same logic couldn't be applied here
$$a_1=1$$
$$a_n=a_{n-1}+2n-1,n\ge2$$
heres my solution 
$$d=2n-1$$
$$a_n=a_1+(n-1)*(2n-1)$$
$$a_n=a_1+2n^2-3n+1$$
$$a_1=1$$
$$a_n=2n^2-3n+2$$
Then i couldn't simplify it futher. But the answer says else-wise
$$a_n=n^2$$

REPLY [2 votes]: One way is to note, that that because n is variable, differences are in arithmetic progression in this case.
This means, we can sum up all the differences as in an arithmetic progression.  $n-1$ terms gives us ${(n-1)(3+(2n-1))\over 2}$ we can then add this to $a_1$ getting $1+(n-1)(n+1)= 1+n^2-1=n^2$