考慮數列７，１０，１３，１６，．．．
回答 (5)
2006-12-04 5:15 am
✔ 最佳答案
a4+3x
2006-12-03 21:16:21 補充：
5(4 3*5) - (4 3*2)= 19 - 10= 9
參考: myself
2006-12-05 3:35 am
第n項=3n+4
第５項=3(5)+4
=19
第６項=3(6)+4
=22
第５項與第６項的和
=19+22
=41
第５項=3(5)+4
=19
第６項=3(6)+4
=22
第５項與第６項的和
=19+22
=41
2006-12-05 3:27 am
a this is+3
b 41
2006-12-24 20:40:19 補充：
a 7+3=10
b 19+22=41
10+3=13
13+3=16
b 41
2006-12-24 20:40:19 補充：
a 7+3=10
b 19+22=41
10+3=13
13+3=16
參考: me
2006-12-04 6:46 am
a)3n+4
b)when n=5
3(5)+4
=19
when n=6
3(6)+4
=22
so sum of 5th term and 6th term
=19+22
=41
b)when n=5
3(5)+4
=19
when n=6
3(6)+4
=22
so sum of 5th term and 6th term
=19+22
=41
2006-12-04 5:48 am
Let d be the difference of two numbers,
n be the no. of terms
Formula : 7 + (n - 1) d
The 1st difference (d) between 7 & 10:
7 + d = 10
d = 3
Proof: 數列的第 2 項,
7 + (n - 1) d
= 7 + (2 - 1) (3)
= 7 + 3
= 10
Therefore, 數列的第５項,
7 + (n - 1) d
= 7 + (5 - 1) (3)
= 7 + (4) (3)
= 19
Therefore, 數列的第 6 項,
7 + (n - 1) d
= 7 + (6 - 1) (3)
= 7 + (5) (3)
= 22
數列的第５項與第６項的和 = 19 + 22
= 41
n be the no. of terms
Formula : 7 + (n - 1) d
The 1st difference (d) between 7 & 10:
7 + d = 10
d = 3
Proof: 數列的第 2 項,
7 + (n - 1) d
= 7 + (2 - 1) (3)
= 7 + 3
= 10
Therefore, 數列的第５項,
7 + (n - 1) d
= 7 + (5 - 1) (3)
= 7 + (4) (3)
= 19
Therefore, 數列的第 6 項,
7 + (n - 1) d
= 7 + (6 - 1) (3)
= 7 + (5) (3)
= 22
數列的第５項與第６項的和 = 19 + 22
= 41
參考: Me
收錄日期: 2021-04-12 22:04:45
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https://hk.answers.yahoo.com/question/index?qid=20061203000051KK05289