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 sum of the previous term f(n-1) and 7.
f(2) = 10 f(5) = 31 f(10) = 66
f(1)&=3 f(n)&=f(n-1)+7, for n>1 To do so, we will use a table.
| n | f(n)=f(n-1)+7 | f(n) |
|---|---|---|
| 1 | f( 1)=3 | 3 |
| 2 | f( 2)=f( 2-1)+7 ⇕ f(2)= f(1)+7 |
f(2)= 3+7 ⇕ f(2)= 10 |
| 3 | f( 3)=f( 3-1)+7 ⇕ f(3)= f(2)+7 |
f(3)= 10+7 ⇕ f(3)= 17 |
| 4 | f( 4)=f( 4-1)+7 ⇕ f(4)= f(3)+7 |
f(4)= 17+7 ⇕ f(4)= 24 |
| 5 | f( 5)=f( 5-1)+7 ⇕ f(5)= f(4)+7 |
f(5)= 24+7 ⇕ f(5)= 31 |
| 6 | f( 6)=f( 6-1)+7 ⇕ f(6)= f(5)+7 |
f(6)= 31+7 ⇕ f(6)= 38 |
| 7 | f( 7)=f( 7-1)+7 ⇕ f(7)= f(6)+7 |
f(7)= 38+7 ⇕ f(7)= 45 |
| 8 | f( 8)=f( 8-1)+7 ⇕ f(8)= f(7)+7 |
f(8)= 45+7 ⇕ f(8)= 52 |
| 9 | f( 9)=f( 9-1)+7 ⇕ f(9)= f(8)+7 |
f(9)= 52+7 ⇕ f(9)= 59 |
| 10 | f( 10)=f( 10-1)+7 ⇕ f(10)= f(9)+7 |
f(10)= 59+7 ⇕ f(10)= 66 |
Therefore, f(2) = 10, f(5) = 31 and f(10) = 66.