Page 159 - NUMINO Challenge_C1
P. 159
Answer
Key
Rules and Numbers 5 The 100th term of the sequence is the 50th term
of Sequence , therefore 50 2 100.
10 Sequence Types p.86~p.87 Problem solving
Example 4 1 Divide the sequence into two sequences such
4, 4, 4, 4, 4; 30, 121; 30
that each sequence only contains the terms of
TryItAgain Each term is obtained by adding a number an odd index or an even index.
that increases by 2 to the previous term. Sequence : 1, 2, 4, 8, 16,
Sequence : 1, 4, 7, 10, 13,
2, 4, 8, 14, 22, 32, 44, In Sequence , each term is obtained by
multiplying the previous term by 2.
2 4 6 8 10 12 1 2 2 2 2 2 2 64, so the index of 64
is 7 in Sequence . Therefore, the index of 64 is
Therefore, the 12th term of this sequence is 2 7 1 13 in the whole sequence.
44 (14 16 18 20 22) 134. In Sequence , each term is obtained by adding
3 to the previous term. 1 3 21 64, so the
Example 16, 48, 17 index of 64 is 22 in Sequence . Therefore, the
17, 18 index of 64 is 2 22 44 in the whole sequence.
18
TryItAgain Each group contains the numbers: (1, 2, 3), 2 The numerators increase by 1. The addends of
(3, 4, 5), (5, 6, 7),
The first number in each group increases by 2. the denominators increase by 1.
In each group, the numbers increase by 1.
Therefore, 20 3 6 r2, and the 20th term of 111111
the sequence is the second number of the 7th
group. The first number of the 7th group is 1234567
1 (2 2 2 2 2 2) 13, therefore the 2 , 3 , 5 , 8 , 12 , 17 , 23 ,
20th term of the sequence is 13 1 14.
123456
Therefore, the numerator of the 14th fraction is
14, and the denominator is
2 (1 2 3 12 13) 93
14
93
Type 10-1 Sequences with Two or More Patterns p.88~p.89 Type 10-2 Sequences with Patterned Groups p.90~p.91
1 6, 7 ; 12, 14 11 121231234
1, 2, 1, 3, 2, 1, 4, 3, 2, 1,
2 The numbers increase by 1 in Sequence .
Therefore, the index of 30 is 30. 23 4 5
3 The numbers increase by 2 in Sequence .
Therefore, the index of 30 is 15.
4 30 2 1 59, 15 2 30
NUMINO Challenge C1
Key
Rules and Numbers 5 The 100th term of the sequence is the 50th term
of Sequence , therefore 50 2 100.
10 Sequence Types p.86~p.87 Problem solving
Example 4 1 Divide the sequence into two sequences such
4, 4, 4, 4, 4; 30, 121; 30
that each sequence only contains the terms of
TryItAgain Each term is obtained by adding a number an odd index or an even index.
that increases by 2 to the previous term. Sequence : 1, 2, 4, 8, 16,
Sequence : 1, 4, 7, 10, 13,
2, 4, 8, 14, 22, 32, 44, In Sequence , each term is obtained by
multiplying the previous term by 2.
2 4 6 8 10 12 1 2 2 2 2 2 2 64, so the index of 64
is 7 in Sequence . Therefore, the index of 64 is
Therefore, the 12th term of this sequence is 2 7 1 13 in the whole sequence.
44 (14 16 18 20 22) 134. In Sequence , each term is obtained by adding
3 to the previous term. 1 3 21 64, so the
Example 16, 48, 17 index of 64 is 22 in Sequence . Therefore, the
17, 18 index of 64 is 2 22 44 in the whole sequence.
18
TryItAgain Each group contains the numbers: (1, 2, 3), 2 The numerators increase by 1. The addends of
(3, 4, 5), (5, 6, 7),
The first number in each group increases by 2. the denominators increase by 1.
In each group, the numbers increase by 1.
Therefore, 20 3 6 r2, and the 20th term of 111111
the sequence is the second number of the 7th
group. The first number of the 7th group is 1234567
1 (2 2 2 2 2 2) 13, therefore the 2 , 3 , 5 , 8 , 12 , 17 , 23 ,
20th term of the sequence is 13 1 14.
123456
Therefore, the numerator of the 14th fraction is
14, and the denominator is
2 (1 2 3 12 13) 93
14
93
Type 10-1 Sequences with Two or More Patterns p.88~p.89 Type 10-2 Sequences with Patterned Groups p.90~p.91
1 6, 7 ; 12, 14 11 121231234
1, 2, 1, 3, 2, 1, 4, 3, 2, 1,
2 The numbers increase by 1 in Sequence .
Therefore, the index of 30 is 30. 23 4 5
3 The numbers increase by 2 in Sequence .
Therefore, the index of 30 is 15.
4 30 2 1 59, 15 2 30
NUMINO Challenge C1
