F.1數學

Each of the following is a number pattern of a sequence,find the first four term of each sequence.
1) 3+n
2) 3n
3) 3n²
4) (3n)²

翻譯後:

以下是序列的數字樣式，找出每個序列的前四期限。
1) 3+n
2) 3n
3) 3n ²
4) (3n) ²

1)3+n

首先,把1代入n.

3+n

=3+(1)

=4

然後,把2代入n.

3+n

=3+(2)

=5
------------------------
如此類推:
------------------------
n=3,

3+n

=3+(3)

=6

---------------------------

n=4,

3+n

3+n

=3+(4)

=7

---------------------------

2)3n

n=1,

3n

=3(1)

=3

------------------------

n=2,

3n

=3(2)

=6

------------------------

n=3

=3(3)

=9

----------------------

n=4

=3(4)

=12

----------------------

3)3n²

n=1,

=3(1²)

=3

-----------------------

n=2,

=3(2²)

=3(4)

=12

----------------------

n=3,

=3(3²)

=3(9)

=27

---------------------

n=4,

3n²

=3(4²)

=3(16)

=48

----------------------------

4)(3n)²

n=1

(3n²)

=[(3)(1)]²

=3²

=9

-----------------------

n=2

(3n²)

=[(3)(2)]²

=6²

=36

--------------------

n=3

(3n)²

=[(3)(3)]²

=9²

=81
-------------------

n=4,

(3n²)

=[(3)(4)]²

=12²

=144

------------------------

1) 4,5,6,7
2) 3,6,9,12
3) 3,12,27,48
4) 9,39,81,144

It is too simple, why bother ask here?

For the first term you put n =1,
the second term you put n =2,
the third term you put n =3,
the fourth term you put n =4,

so
1) first term = 3 +1 = 4
second term = 3+2 = 5
third term = 6
fourth term = 7

2) first term, second term, third term, fourth term
are 3, 6, 9, 12

3) first term, second term, third term, fourth term
are 3*1, 3*2^2 , 3* 3^2, 3*4^2 = 3, 12, 27, 48

4) first term, second term, third term, fourth term
are (3*1)^2 , (3*2)^2, (3*3)^2, (3*4)^2
= 9, 36, 81, 144