smc

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, specifically, the one that contains the points $(1,2),(2,4),$, and $(3,8)$. Let the quadratic be $f(x)=a{x}^2+bx+c$, 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 $3a+b=2$ (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 $a_{100}=100^2-100+2=9902,\boxed{9902}$.