The equation for this sequence is:

a_n = a_n-1 + 3

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 3
a₆ = 13 + 3
a₆ = 16

To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:

\( \sqrt{\frac{49}{81}} \)
\( \frac{\sqrt{49}}{\sqrt{81}} \)
\( \frac{\sqrt{7^2}}{\sqrt{9^2}} \)
\(\frac{7}{9}\)

To add these radicals together their radicands must be the same:

7\( \sqrt{20} \) + 3\( \sqrt{5} \)
7\( \sqrt{4 \times 5} \) + 3\( \sqrt{5} \)
7\( \sqrt{2^2 \times 5} \) + 3\( \sqrt{5} \)
(7)(2)\( \sqrt{5} \) + 3\( \sqrt{5} \)
14\( \sqrt{5} \) + 3\( \sqrt{5} \)

Now that the radicands are identical, you can add them together:

The equation for this sequence is:

a_n = a_n-1 + 3(n - 1)

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 3(6 - 1)
a₆ = 31 + 3(5)
a₆ = 46

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{6!}{3!} \)
\( \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{6 \times 5 \times 4}{1} \)
\( 6 \times 5 \times 4 \)
120