Remainders

We have that $a_{n} = (7 - 1)^{n} + (7 + 1)^{n}$ . Using the binomial formula, we find that

$a_{n} = \displaystyle\sum_{k=0}^{n}\dbinom{n}{k}7^{n-k}(-1)^{k} + \sum_{k=0}^{n}\dbinom{n}{k}7^{n-k}(1)^{k} = \sum_{k=0}^{n}\dbinom{n}{k}7^{n-k}((-1)^{k} + 1^{k})$ .

All the terms where $k$ is odd will then cancel out, while for all the terms where $k$ is even we will end up multiplying that term by $2$ . Thus those terms with $n - k \ge 2$ will have a factor of $2*7^{2} = 98$ , which mod $98$ will disappear, leaving us to simply consider the term $n - k = 1$ if $n$ is odd, (since only even $k$ are still in play), or where $n - k = 0$ if $n$ is even. In this case, we thus only need consider the term where $n = k = 98$ , giving us that

$a_{98} \pmod{98} = \dbinom{98}{98}*7^{0}*2 = \boxed{2}$ .