Arrange the continuous natural numbers 1 to 1001 into a rectangular array as shown in the figure, and frame 16 numbers with a square, Is it possible to make the sum of the 16 numbers in this square equal to: (1) 1988, (2) 1991, (3) 2000, (4) 2080? If not, try to explain the reason. If possible, please write down the minimum and maximum of the 16 numbers in this square one 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ... ... 995 996 997 998 999 1000 1001 1 2 3 4 5 6 7

These 16 numbers are respectively n n + 1 N + 2 N + 3 N + 7 n + 8 N + 9 n + 10 N + 14 n + 15 N + 16 n + 17 n + 21 n + 22 n + 23 n + 2416, the sum is 16N + 1921.16n + 192 = 1988, n = 112.25, n is not an integer, 2.16n + 192 = 1991, n = 112.4375, n is not an integer, and 1991 is an odd number, so we can round 3

Arrange the natural numbers from small to large and try to find the sum of the first n terms
Add a question and give 20 points for the answer
Given 2x + 3Y = 0, find the algebraic formula ① 5x + 4Y divided by 3x-2y, ② X & # 178; + xy-y & # 178; divided by X & # 178; - XY + Y & # 178;

n*(n-1)/2

The sum of seven consecutive natural numbers is 63. How many are the seven natural numbers arranged from small to large?

63 / 7 = 9, the number in the middle is 9
These seven natural numbers are arranged from small to large
6 7 8 9 10 11 12

There are six natural numbers (arranged from small to large). The difference between every two adjacent numbers is 4. If the first number is a, what is the sum of these six natural numbers?

a+a+4+a+8+a+12+a+16+a+20
=6a+60

In the array of natural numbers, what is the natural number directly below 1000? 1 　　2 　　5 　　… 4 　　3 　　6 　　… 9 　　8 　　7 　　……

∵ the number of the first column is & nbsp; 12,22,32 961 = 312 ＜ 1000 ＜ 322 = 1024, the number of 1024-1000 + 1 = 25 from the left of line 32, and the number of 332-25 + 1 = 1065 from the left of line 33 is directly below the number of 1024-1000 + 1 = 25

（1/2+1/3+...+1/2004)(1+1/2+...+1/2003）-（1+1/2+..+1/2004）（1/2+1/3+..+1/2003）

Let x = 1 / 2 + 1 / 3 +... + 1 / 2004
y=1/2+...+1/2003
（1/2+1/3+...+1/2004)(1+1/2+...+1/2003）-（1+1/2+..+1/2004）（1/2+1/3+..+1/2003）
=x(1+y)-(1+x)y
=x+xy-y-xy
=x-y
=1/2004

Using decomposition factor to calculate ① 3 ^ 2004-3 ^ 2003 ② (- 2) ^ 101 + (- 2) ^ 100 3Q!

①3^2004-3^2003 ＝3^2003(3-1)=2*3^2003
②（-2）^101+(-2)^100 =(-2)^100(-2+1)=－2^100

(1/2004-1)(1/2003-1)(1/2003-1)… (1/3-1)(1/2-1)
Better be before eight

(1/2004-1)(1/2003-1)(1/2002-1)… (1/3-1)(1/2-1)
=(- 2003 / 2004) × (- 2002 / 2003) × (- 2001 / 2002) × (- 2 / 3) × (- 1 / 2) (there are 2003 items, which are odd)
=-1/2004

111.. 1 (2004 1) - 222.. 2 (1002 2 2) is equal to a × a to find a

111.. 1 (2004 1) - 222.. 2 (1002 2 2)
=111.. 1 (2004 1) - 111.. 1 (1002 1) - 111.. 1 (1002 1)
=11... 100... 0 (1002 11002 0) - 111.. 1 (1002 1)
=111.. 1 (1002 zeros) * (100... 0 (1002 zeros) - 1)
=111.. 1 (1002 pieces of 1) * 99... 9 (1002 pieces of 9)
=111.. 1 ^ 2 (1002 ones) * 9
=33... 3 ^ 2 (1002 3)
So a = 333... 3 (1002 3)

Calculate 111 12004 1-222 21002 2 = a × a, find a

Because 11-2 = 3 × 31111-22 = 33 × 33111111-222 = 333 × 333 12004 1-222 21002 2 =, so a = 333 3 (1002 3)

Arrange the natural numbers from small to large and try to find the sum of the first n terms Add a question and give 20 points for the answer Given 2x + 3Y = 0, find the algebraic formula ① 5x + 4Y divided by 3x-2y, ② X & # 178; + xy-y & # 178; divided by X & # 178; - XY + Y & # 178;