Mathematical pyramid According to the law of 123456 What is the mth number in row n?

The trick is to look at the pattern of the last number in each row.

It is not difficult to find that the last number in the nth row = 1+2+3+.n=(1 2)n(1+n) (formula).

So we know the last number in the nth row.

Then the mth number is the last number in the row - (n-m).

So the mth number in row n = (1 2)n(1+n)-(n-m) = (1 2)n 2-1 2n+m.

The square of one-half n minus one-half n plus m.

Stipulate. 1, 2, 4, 7, 11, 16, (22), (29), the difference is: 1, 2, 3, 4, 5, 6 ,...

2, 5, 10, 17, 26, (37), (50), the difference is: 3, 5, 7, 9 ,...

0, 3, 8, 15, 24, (35), (48), - the difference is: 3, 5, 7, 9 ,...

Find the pattern and fill in the blanks: 9-1=8, 16-4=12, 25-9=16, 36-16=20, 49-25=24.

Don't know that your pyramid is platoons. If I estimated it.

Words like this. The trick is to look at the pattern of the last number in each row.

It is not difficult to find that the last number in the nth row = 1+2+3+.n=(1 2)n(1+n) (formula).

So we know the last number in the nth row.

Then the m number is the last number of the row - (n-m), so the m number of the n row = (1 2) n(1+n) - (n-m) = (1 2) n 2-1 2n + m

The square of one-half n minus one-half n plus m.

The law of increment is that the next number is 1 more than the previous number, all the way to 6

Third line 12 20

The fourth row 32 is the addition of two adjacent numbers. 1357 1+3=4；3+5=8；5+7=12

A: Strictly speaking, this is not a very regular pyramid, it is just borrowed from the "tower" method.

First line 1357

Second line 4812

Third line 1293

Fourth line 31112

The first number of the nth layer is (n-1) squared.

The square of 44 (the 1st of the 44th layer) is 1936, the square of 45 (the 1st of the 45th layer) is 2025, and you can deduce that 2015 is the number of the 44th line.

1 number in 1 row, 2 numbers in 2 rows, n number in n rows, 1+2+3+...n numbers, this sum is also the last number in the nth row, so what is needed is to see how much n is when its sum is around 2007; with formulas.

n*(n+1) 2=2007 solution n=;

62*(62+1) 2=1953 < 2007, i.e. the last number in line 62 is 1953;

63*(63+1) 2=2016 > 2007, i.e. the last number in line 63 is 2016;

So 2007 is in line 63, and numbers 2007-1922 = 54.

From the first line to the nth row, there are a total of:

1+2+3+..n = (1 + n) n 2 numbers.

Since 1+2+...63=(1+63)*63/2=20161+2+..62=(1+62)*62 2=19532007 between 1953 and 2016, so 2007 should be in line 62+1=63.

2007-1953 = 54 columns.

Each row has one more number in turn, and there are n numbers in the nth row, so to find it, you can use (1+n)*n 2 to find the number.

At the end of line 62 is 1953, and to the next line 2007-1953=54, so it is line 63 number 54.

What is this pattern?

There are 2n-1 numbers in the nth row, so there are 2000 rows in total (1+3+5+......3999) = (1 + 3999) * 2000 2 = 4000000 The 2001st number in line 2002 should be 4002002