Question about arithmetic progression. stuck on one of the answers 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
$$a_n=n^2$$
 A: Hint:
Remember that an arithmetic progression is a series where two consecutive terms have the same common difference.
Also, note that $a_n,a_{n+1}$ are two consecutive terms and $a_{n-1},a_n$ are two consecutive terms.
So, by definition, $a_n-a_{n-1}$ and $a_{n+1}-a_n$ should be constant common difference.
However, for the second question, the common difference is dependent on $n$ and thus not constant. So the second one is not an AP.
Hoewvwer an alternate approach to solve the second problem exists.
Using the fact $a_n-a_{n-1}=2n-1$ we get,
$$a_2-a_1=3$$
$$a_3-a_2=5$$
$$a_4-a_3=7$$
$$.$$
$$.$$
$$a_n-a_{n-1}=2n-1$$
Adding the two columns we have 
$$a_n-a_1=3+5+7+...+(2n-1)$$
$$\implies a_n=1+3+5+7+...+(2n-1)\text{ [since a_1=1]}$$
So now the right side is an AP. 
Add it and voila!
A: 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 $1+(n-1)(n+1)= 1+n^2-1=n^2$
A: You can't treat the second recurrence as an arithmetic progression since its step size isn't constant.
That $a_n=n^2$ there may perhaps be most easily seen visually:
...o... a_1
..oox.. a_2
.oooxx. a_3
ooooxxx a_4

and then you fold the triangle of x onto that of o, obtaining a square of size $n$. The formal proof may be achieved via induction.
