The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule
. The next number would then be fourth power of 7 plus 1, or 2402.
And the harder way: Denote the <em>n</em>-th term in this sequence by
, and denote the given sequence by
.
Let
denote the <em>n</em>-th term in the sequence of forward differences of
, defined by

for <em>n</em> ≥ 1. That is,
is the sequence with





and so on.
Next, let
denote the <em>n</em>-th term of the differences of
, i.e. for <em>n</em> ≥ 1,

so that




etc.
Again: let
denote the <em>n</em>-th difference of
:




etc.
One more time: let
denote the <em>n</em>-th difference of
:



etc.
The fact that these last differences are constant is a good sign that
for all <em>n</em> ≥ 1. Assuming this, we would see that
is an arithmetic sequence given recursively by

and we can easily find the explicit rule:



and so on, up to


Use the same strategy to find a closed form for
, then for
, and finally
.




and so on, up to

Recall the formula for the sum of consecutive integers:







and so on, up to

Recall the formula for the sum of squares of consecutive integers:










