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Emergency treatment of several math problems with regularity found in sixth grade
1、

The first number is 1= 1+0.

The second number is 2= 1+ 1.

The third number is 4= 1+ 1+2.

The fourth number is 7= 1+ 1+2+3.

The fifth number is11=1+1+2+3+4.

The sixth number is16 =1+1+2+3+4+5.

The seventh number is 22= 1+ 1+2+3+4+5+6.

The eighth number is 29 =1+1+2+3+4+5+6+7.

What number is 1994? = 1+ 1+2+3+4+5+……+ 1993= 1+ 1993*( 1993+ 1)/2 = 1987022

So the remainder is 2.

2、

3、

You can divide the data into two columns: the original 1, 3,5, ... to the left and the original 2,4,6, ... to the right.

2 4 6 8 | 10 12 14 16

18 20 22 24 | 26 28 30 32

34 36 38 40 | 42 44 46 48

According to the law, this series starts from even numbers, so (1996/2)/8= 124+6.

Therefore, the position of 1996 in the new arrangement is 124, the sixth, so the line that restores the original arrangement is 124*2+ 1=249, and the position is 6-4=2, so it is the second of 249 lines.

4. The last number in the n line is n 2,15120 =122 2+236.

So 15 120 is the 236th position in line 123.