Question

Simplify the following:

`(3^(n + 4) xx 3^((n - 2)(n + 2)))/(3^(n(n + 1)) xx 9^(n + 1))`

Answer 1

Given: `(3^(n + 4) xx 3^((n - 2)(n + 2)))/(3^(n(n + 1)) xx 9^(n + 1))`

Step-wise calculation:

1. Combine powers of 3 in the numerator:

`3^(n + 4) xx 3^((n - 2)(n + 2)) = 3^((n + 4 + (n - 2)(n + 2))`

Use difference of squares on (n – 2)(n + 2):

(n – 2)(n + 2) = n² – 4

So the numerator exponent is:

n + 4 + n² – 4 = n² + n

Therefore, the numerator is:

`3^(n^2 + n)`

2. The denominator is:

`3^(n(n + 1)) xx 9^(n + 1)`

Note that 9 = 3², so:

`9^(n + 1) = (3^2)^(n + 1)`

`9^(n + 1) = 3^(2(n + 1))`

`9^(n + 1) = 3^(2n + 2)`

Therefore, the denominator can be rewritten as:

`3^(n^2 + n) xx 3^(2n + 2) = 3^((n^2 + n) + (2n + 2))`

`3^(n^2 + n) xx 3^(2n + 2) = 3^(n^2 + 3n + 2)`

3. Now, the full expression becomes:

`(3^(n^2 + n))/(3^(n^2 + 3n + 2)) = 3^((n^2 + n) - (n^2 + 3n + 2))`

`(3^(n^2 + n))/(3^(n^2 + 3n + 2)) = 3^(-2n - 2)`