Answer
a) 2, 5, 8, 11, 14, 17, 20, 23, 26, 29
b) 1, 1, 1, 2, 2, 2, 3, 3, 3, 4
c) 1, 1, 3, 3, 5, 5, 7, 7, 9, 9
d) -1 ,-2, -2, 8, 88, 656, 4912, 40064, 362368, 3627776
e) 3 ,6, 12, 24, 48, 96, 192, 384, 768, 1536
f) 2, 4, 6, 10, 16, 26, 42, 68, 110, 178
g) 1, 2, 2, 3, 3, 3, 3, 4, 4, 4
h) 3, 3, 5, 4, 4, 3, 5, 5, 4, 3
Work Step by Step
A.
Sequence starts with 2, thus a1 = 2.
Each next term is the previous term that will be increased by 3.
a2 = a1 + 3 = 2 + 3 = 5
a3 = a2 + 3 = 5 + 3 = 8
a4 = a3 + 3 = 8 + 3 = 11
a5 = a4 + 3 = 11 + 3 = 14
a6 = a5 + 3 = 14 + 3 = 17
aA7 = a6 + 3 = 17 + 3 = 20
a8 = a7 + 3 = 20 + 3 = 23
a9 = a8 + 3 = 23 + 3 = 26
a10 = a9 + 3 = 26 + 3 = 29
B.
Sequence involves positive numbers in increasing numbers starting from 1 as the first term. Each term happens 3 times.
a1 = 1
a2 = 1
a3 = 1
a4 = 2
a5 = 2
a6 = 2
a7 = 3
a8 = 3
a9 = 3
a10 = 4
C.
Sequence involves odd positive numbers in increasing numbers starting from 1 as the first term. Each term happens twice.
a1 = 1
a2 = 1
a3 = 3
a4 = 3
a5 = 5
a6 = 5
a7 = 7
a8 = 7
a9 = 9
a10 = 9
D.
Using the given an=n!−2^n , substitute and evaluate An at n=1,2,3,4,5,6,7,8,9,10.
a1 = 1! -2^1
a1 = 1 - 2 = -1
a2 = 2! -2^2
a2 = 2 - 4 = -2
a3 = 3! -2^3
a3 = 6 - 8 = -2
a4 = 4! -2^4
a4 = 24 - 16 = 8
a5 = 5! -2^5
a5 = 120 - 32 = 88
a6 = 6! -2^6
a6 = 720 - 64 = 656
a7 = 7! -2^7
a7 = 5040 - 128 = 4912
a8 = 8! -2^8
a8 = 40320 - 256 = 40064
a9 = 9! -2^9
a9 = 362880 - 512 = 362368
a10 = 10! -2^10
a10 = 3628800 - 1024 = 3627776
E.
Sequence begins with the first term = 3.
a1 = 3
Each term is twice the previous term.
a2 = 2a1 = 2(3) = 6
a3 = 2a2 = 2(6) = 12
a4 = 2a3 = 2(12) = 24
a5 = 2a4 = 2(24) = 48
a6 = 2a5 = 2(48) = 96
a7 = 2a6 = 2(96) = 192
a8 = 2a7 = 2(192) = 384
a9 = 2a8 = 2(384) = 758
a10 = 2a9 = 2(768) = 1536
F.
First term = 2.
Second term = 4.
Every next term is the sum of the two previous term.
a3 = 2 + 4 = 6
a4 = 4 + 6 = 10
a5 = 6 + 10 = 16
a6 = 10 + 16 = 26
a7 = 16 + 26 = 42
a7 = 26 + 42 = 68
a9 = 42 + 68 = 110
a10 = 68 + 110 = 178
G.
The “n” term is the number of bits in the binary expansions of the number(n).
Binary expansion of 1 = 1
a1 = 1
Binary expansion of 2 = 10
a2 = 2
Binary expansion of 3 = 11
a3 = 2
Binary expansion of 4 = 100
a4 = 3
Binary expansion of 5 = 101
a5 = 3
Binary expansion of 6 = 110
a6 = 3
Binary expansion of 7 = 111
a7 = 3
Binary expansion of 8 = 1000
a8 = 4
Binary expansion of 9 = 1001
a9 = 4
Binary expansion of 10 = 1010
a10 = 4
H.
The “n” term is the number of letters in the word n.
a1 = 3 (one)
a2 = 3 (two)
a3 = 5 (three)
a4 = 4 (four)
a5 = 4 (five)
a6 = 3 (six)
a7 = 5 (seven)
a8 = 5 (eight)
a9 = 4 (nine)
a10 = 3 (ten)
