If the [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$.
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$
We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern:
$2, 4, 8, 14, 22, ....$. We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$, a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: each term is the previous term plus the next even number. Therefore, since the differences of consecutive terms form an arithmetic sequence, then the terms satisfy a [quadratic](https://artofproblemsolving.com/wiki/index.php/Quadratic), specifically, the one that contains the points $(1, 2), (2, 4),$, and $(3, 8)$. Let the quadratic be $f(x)=ax^2+bx+c$, so:
$a+b+c=2$ (1)
$4a+2b+c=4$ (2)
$9a+3b+c=8$ (3)
Subtracting (1) from (2) and (2) from (3) yields the two-variable [system of equations](https://artofproblemsolving.com/wiki/index.php/System_of_equations)
$3a+b=2$ (4)
$5a+b=4$ (5)
We can subtract (4) from (5) to find that $2a=2$, so $a=1$. Substituting this back in yields $b=-1$, and substituting these back into one of the original equations yields $c=2$, so the closed form for the terms is
$f(x)=x^2-x+2$, or
$a_n=n^2-n+2$.
Substituting in $n=100$ yields $a_{100}=100^2-100+2=9902, \boxed{\text{B}}$.