smc

Using the recursion from solution 1, we see that the first differences of $4,8,12,...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can be generated from a quadratic function. If $f(n)=a{n}^2+bn+c$, and $f(0)=1$, $f(1)=5$, and $f(2)=13$, we get a system of three equations in three variables: $f(0)=1$ gives $c=1$ $f(1)=5$ gives $a+b+c=5$ $f(2)=13$ gives $4a+2b+c=13$ Plugging in $c=1$ into the last two equations gives $a+b=4$ $4a+2b=12$ Dividing the second equation by 2 gives the system: $a+b=4$ $2a+b=6$ Subtracting the first equation from the second gives $a=2$, and hence $b=2$. Thus, our quadratic function is: $f(n)=2n^2+2n+1$ Calculating the answer to our problem, $f(100)=20000+200+1=20201$, which is choice $\boxed{20201}$.