Question

For each recursively-defined sequence below, write the first few term. Then use the terms to write an explicit equation.

Answer 1

Answer:

Step-by-step explanation:

a_1 = 15

a_2 = a_1  - 3

a_2 = 15 - 3

a_2 = 12

a_3 = a_2  - 3

a_3 = 12 - 3

a_3 = 9

a_4 = a_3 - 3

a_4 = 9 - 3

a_4 = 6

a_5 = 3

a_6 = 0                  I'm leaving these last two to expand

a_n = a1 - (n - 1)*d

a_n = 15 - (n - 1)*3

a_n = 15 - 3n + 3

a_n = 18 - 3n

Example

a_6 = 18 - 3*6

a_6 = 0

Problem B

t(1) = 108

t(1 + 1) = 1/3 * 108

t(2) = 36

t(3) = 1/3 * t2

t(3) = 1/3 * 36

t(3) = 12

t(4) =1/3 (t(3))

t(4) = 1/3 * 12

t(4) = 4

t(5) = t4 / 3

t(5) = 4 / 3

t(5) = 1.3333333

So the explicit definition is

t(n) = 108 (1/3)^(n - 1)            You could simplify this a little bit by realizing that 108 is made of three 3s.

t(n) = 4 * 3^3 * (1/3)^(n - 1)

t(n) = 4 * (1/3) ^ (n - 4)

Example

t(5) = 108 (1/3)^4

t(5) = 108(1/81)

t(5) = 1.3333333

And using the simplified formula, you get.

t(5) = 4 * (1/3)^1

t(5) = 1.333333 which is the same thing as the original result.

Answer 2

Answer:

see explanation

Step-by-step explanation:

(a)

Using the recursive rule and a₁ = 15 , then

a₂ = a₁ - 3 = 15 - 3 = 12

a₃ = a₂ - 3 = 12 - 3 = 9

a₄ = a₃ - 3 = 9 - 3 = 6

The sequence is 15, 12, 9, 6, ....

This is an arithmetic sequence with n th term ( explicit rule )

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 15 and d = - 3 , then

$a_{n}$ = 15 - 3(n - 1) = 15 - 3n + 3 = 18 - 3n

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(b)

Using the recursive rule and t₁ = 108 , then

t₂ = $\frac{1}{3}$ t₁ = $\frac{1}{3}$ × 108 = 36

t₃ = $\frac{1}{3}$ t₂ = $\frac{1}{3}$ × 36 = 12

t₄ = $\frac{1}{3}$ t₃ = $\frac{1}{3}$ × 12 = 4

The sequence is 108, 36, 12, 4, ......

This is a geometric sequence with n th term

$t_{n}$ = t₁$(r)^{n-1}$

where t₁ is the first term and r the common ratio

Here t₁ = 108 and r = $\frac{1}{3}$ , then

$t_{n}$ = 108 $(\frac{1}{3}) ^{n-1}$