6. Recursively Defined Sequences
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The given rule means that, after the first term of the sequence, every term f(n) is the product of the previous term f(n-1) and 6.
f(2) = - 6 f(5) = - 1296 f(10) = - 10 077 696
f(1)&=- 1 f(n)&=6f(n-1), for n>1 To do so, we will use a table.
| n | f(n)=6f(n-1) | f(n) |
|---|---|---|
| 1 | f( 1)=- 1 | - 1 |
| 2 | f( 2)=6f( 2-1) ⇕ f(2)=6 f(1) |
f(2)=6( - 1) ⇕ f(2)= - 6 |
| 3 | f( 3)=6f( 3-1) ⇕ f(3)=6 f(2) |
f(3)=6( - 6) ⇕ f(3)= - 36 |
| 4 | f( 4)=6f( 4-1) ⇕ f(4)=6 f(3) |
f(4)=6( - 36) ⇕ f(4)= - 216 |
| 5 | f( 5)=6f( 5-1) ⇕ f(5)=6 f(4) |
f(5)=6( - 216) ⇕ f(5)= - 1296 |
| 6 | f( 6)=6f( 6-1) ⇕ f(6)=6 f(5) |
f(6)=6( - 1296) ⇕ f(6) - 7776 |
| 7 | f( 7)=6f( 7-1) ⇕ f(7)=6 f(6) |
f(7)=6( - 7776) ⇕ f(7)= - 46 656 |
| 8 | f( 8)=6f( 8-1) ⇕ f(8)=6 f(7) |
f(8)=6( - 46 656) ⇕ f(8)= - 279 936 |
| 9 | f( 9)=6f( 9-1) ⇕ f(9)=6 f(8) |
f(9)=6( - 279 936) ⇕ f(9)= - 1 679 616 |
| 10 | f( 10)=6f( 10-1) ⇕ f(10)=6 f(9) |
f(10)=6( - 1 679 616) ⇕ f(10)= - 10 077 696 |
Therefore, f(2) = - 6, f(5) = - 1296 and f(10) = - 10 077 696.