<s>[INST] 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:
$$A=2452,\:B=2450,\:C=2552,\:D=2500$$
 [/INST] 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):
$$a_{50}-a_1=2(1+2+3+\cdots +49)  \Rightarrow \\
a_{50}=a_1+2\cdot \frac{1+49}{2}\cdot 49=2+2450=2452.$$
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