Best Answer

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

More answers

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

Q: What is the rule for 1 2 6 24?

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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

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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.

Related questions

t(n) = 6*n where n = 1, 2, 3, etc

One possible answer: Un = (28n3 - 207n2 + 459n - 268)/2 for n = 1, 2, 3, ...

t(n) = 24/2n, n = 1, 2, 3, ...

t(n) = (5n4 - 10n3 + 55n2 - 50n + 24)/24 where n = 1, 2, 3, ...

ni = ni-1+6

(7 + 1) × 6 ÷ 2 = 24.

2*6*(-1+3) = 24

(2*9)+(6*1)=24

(9 - 1) * 6 / 2 = 24.

2 6/8 - 5/6 = 2 18/24 - 20/24 = 1 24/24 + 18/24 = 1 42/24 - 20/24 = 1 22/24 or 1 11/12

6 * [5-(2-1)] = 6*[5-1] = 6*4 = 24

(2 + 6) * 1 * 3 = 24.