The document provides three examples of writing the first five terms of sequences defined by different recurrence relations. The first sequence is defined by an=an-2-an-1 and has terms 2, 2, 0, 2, -2. The second sequence is defined by an=an-1+n and has terms 1, 2, 4, 7, 11. The third sequence is defined by a1=4, an=2an-1-1 and has terms 4, 7, 13, 25, 49.

1. EXAMPLE: WRITE THE 1EXAMPLE: WRITE THE 1STST
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TERMS OF THE SEQUENCE.TERMS OF THE SEQUENCE.
•a1=2, a2=2, an=an-2-an-1
a3=a3-2-a3-1→a1-a2=2-2=0
a4=a4-2-a4-1→a2-a3=2-0=2
a5=a5-2-a5-1→a3-a4=0-2=-2
2, 2, 0, 2, -2
11stst
termterm 22ndnd
termterm 1
2
3
4
5
2
2
0
2
-2

2. Write the first five terms of the sequence.Write the first five terms of the sequence.
1.1. aa11 = 3,= 3, aann == aann – 1– 1 77––
SOLUTIONSOLUTION
aa11 = 3= 3
aa33 == aa22 –– 77
aa33 == –– 44 –– 7 =7 = –– 1111
aa44 == aa33 –– 7 =7 = –– 1111 –– 7 =7 = –– 1818
aa55 == aa44 –– 7 =7 = –– 1818 –– 7 =7 = –– 2525
aa22 == aa11 7 = 3 7 = 47 = 3 7 = 4–– ––––
ANSWERANSWER 3, –4, –11, –18, –253, –4, –11, –18, –25
33
−−44
−−1111
−−1818
−−2525

3. Write the first five terms of the sequence.Write the first five terms of the sequence.
2.2. aa00 = 1,= 1, aann == aann – 1– 1 ++ nn
SOLUTIONSOLUTION
aa00 = 1= 1
aa11 == aa00 ++ 1 = 11 = 1 ++ 1 =1 = 22
aa22 == aa11 ++ 1 =1 = 22 ++ 2 = 42 = 4
aa33 == aa22 ++ 3 = 43 = 4 ++ 3 = 73 = 7
aa44 == aa33 ++ 4 = 74 = 7 ++ 4 = 114 = 11
ANSWERANSWER 1, 2, 4, 7, 111, 2, 4, 7, 11

4. Write the first five terms of the sequence.Write the first five terms of the sequence.
3.3. aa11 = 4,= 4, aann = 2= 2aann – 1– 1 –– 11
SOLUTIONSOLUTION
aa11 = 4= 4
aa22 = 2= 2aa11 –– 1 = (2 4)1 = (2 4) –– 1 = 81 = 8 –– 1 = 71 = 7
aa33 = 2= 2aa22 –– 1 = (2 7)1 = (2 7) –– 1 = 141 = 14 –– 1 = 131 = 13
aa55 = 2= 2aa44 –– 1 = (2 25)1 = (2 25) –– 1 = 491 = 49
aa44 = 2= 2aa33 –– 1 = (2 13)1 = (2 13) –– 1 = 261 = 26 –– 1 = 251 = 25
ANSWERANSWER 4, 7, 13 25, 494, 7, 13 25, 49