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(200) High scores are rewarded! ! ! ! Six super difficult math problems.
① 1,2,3,4,(7),6, 1 1

1+2=3

2+3=5

3+4=7

4+7= 1 1

7+6= 13

6+ 1 1= 17

The sum of the two terms is a prime number, arranged in order of size.

② 0,0,6,24,60, 120(2 10 or 336)

Method 1:

0-0=0

6-0=6

24-6= 18

60-24=36

120-60=60

From 0, 6, 18, 36, 60,

We know: 6× 0, 6× 1, 6× (1+2), 6× (1+2+3), 6× (1+2+4),

The difference of the latter term should be 6×( 1+2+3+4+5)=90.

So x- 120=90.

Find x=2 10.

Method 2:

0=0? -0

0= 1? - 1

6=2? -2

24=3? -3

60=4? -4

120=5? -5

2 10=6? -6

Know the general formula: n? Tong -EN

Method 3:

(- 1)×0× 1=0

0× 1×2=0

1×2×3=6

2×3×4=24

3×4×5=60

4×5×6= 120

5×6×7=2 10

6×7×8=336

So the answer to this question is: 0, 0, 6, 24, 60, 120 (2 10 or 336).

③-2,- 14,-24,4, 130,438,( 1036)

1^3×(-3)+ 1=-2

2^3×(-2)+2=- 14

3^3×(- 1)+3=-24

4^3×0+4=4

5^3× 1+5= 130

6^3×2+6=438

7^3×3+7= 1036

④5,67,37,5 15, 10 1( 173 1)

2^2+ 1=5

4^3+3=67

6^2+ 1=37

8^3+3=5 15

10^2+ 1= 10 1

12^3+3= 173 1

⑤ 14,20,54,76,( 126)

3^2+5= 14

5^2-5=20

7^2+5=54

9^2-5=76

1 1^2+5= 126

⑥2,3,2,4,(5),(4),2,7

(2-3)^2+ 1=2

(3-2)^2+3=4

(2-4)^2+ 1=5

(4-5)^2+3=4

(5-4)^2+ 1=2

(4-2)^2+3=7

From the third term, we can see that the latter term is the square of the difference between the first two terms (odd term+1, even term +3), which shows its law.

Honey, do you understand?