There are infinitely many possible answers.
One such is Un = (11n3 - 57n2 + 100n - 48)/6 for n = 1, 2, 3, ...
A more likely one though, is Un = n! = n*(n-1)*...*2*1
There are infinitely many possible answers.
One such is Un = (11n3 - 57n2 + 100n - 48)/6 for n = 1, 2, 3, ...
A more likely one though, is Un = n! = n*(n-1)*...*2*1
There are infinitely many possible answers.
One such is Un = (11n3 - 57n2 + 100n - 48)/6 for n = 1, 2, 3, ...
A more likely one though, is Un = n! = n*(n-1)*...*2*1
There are infinitely many possible answers.
One such is Un = (11n3 - 57n2 + 100n - 48)/6 for n = 1, 2, 3, ...
A more likely one though, is Un = n! = n*(n-1)*...*2*1
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There are infinitely many possible answers.
One such is Un = (11n3 - 57n2 + 100n - 48)/6 for n = 1, 2, 3, ...
A more likely one though, is Un = n! = n*(n-1)*...*2*1
t(n) = 24/2n, n = 1, 2, 3, ...
Common factors of 18 and 24 are: 1, 2, 3, and 6.
The factors of 6 are: 1, 2, 3, 6 The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
-1298
So pretty much it is the same rule as 4 and 6 because both # go into 24. 4 rule: look at the last to places to see if the # is divisible by 4. 6 rule: if the #is both divisible by 2 and 3 then it s divisible by 6.