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.
Summing up the expressions $a_{n+1} = a_n + 2n$ for $n = 1, \dots, k$ one easily gets that $$ a_{k+1} = a_1 + \sum_{i=1}^k 2i = 2 + k(k+1) $$
Hence, $a_{50} = 2 + 2450 = 2452$ and the correct answer is A.
Alternatively, note: $a_{n+1}-a_n=2n$. So: $$\begin{align} a_2-a_1&=2\cdot 1\\ a_3-a_2&=2\cdot 2\\ a_4-a_3&=2\cdot 3\\ &\vdots \\ a_{50}-a_{49}&=2\cdot 49 \end{align}$$ Summing all (midterms telescope): $$a_{50}-a_1=2(1+2+3+\cdots +49) \Rightarrow \\ a_{50}=a_1+2\cdot \frac{1+49}{2}\cdot 49=2+2450=2452.$$
