Arithmetic series problem Given $\left\{a_n\right\}$ arithmetic progression, $a_1=2$, $a_{n+1}=a_n+2n$ $\left(n\:\ge \:1\right)$. $a_{50}=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_{50}=2+d\left(n-1\right)$$
$$a_{50}=2+2\left(n^2-n\right)$$
$$a_{50}=2+2\cdot 2450$$
$$a_{50}=4902$$
But this is wrong. Answers:
$$A=2452,\:B=2450,\:C=2552,\:D=2500$$

\begin{align}a_1&=2\\a_{n+1}&=a_n+2n\\&=a_{n-1}+2[n+(n-1)]\\&=a_{n-2}+2[n+(n-1)+(n-2)]\\
&= \ldots\\
&=a_1+2[n+(n-1)+\ldots +1]\end{align}
Hence \begin{align}a_{50}&=a_1+2(49+\ldots + 1)\\&=2+2\cdot \frac{49(50)}{2}\\&=2+49(50)\\&=2+(50-1)(50)\\&=2+2500-50\\&=2452 \end{align}
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
