(2*3+4*6+6*9+8*12+10*15)/(3*6+6*12+9*18+12*24+15*30) (2 * 3 + 4 * 6 + 6 * 9 + 8 * 12 + 10 * 15) molecule -------------------------------------------= fractional line (3 * 6 + 6 * 12 + 9 * 18 + 12 * 24 + 15 * 30) denominator

(2*3+4*6+6*9+8*12+10*15)/(3*6+6*12+9*18+12*24+15*30) (2 * 3 + 4 * 6 + 6 * 9 + 8 * 12 + 10 * 15) molecule -------------------------------------------= fractional line (3 * 6 + 6 * 12 + 9 * 18 + 12 * 24 + 15 * 30) denominator

Note that the ratio of each item is 1 / 3
So the ratio of their and is also 1 / 3

3+6-9+12+15-18+21+24-27+...+300+303-306

5049
The sum of the first n terms of the arithmetic sequence:
Sn=[n(A1+An)]/2
Sn=nA1+[n(n-1)d]/2
Consider 3 + 6 + 9 + 12 + 15 +... + 306 first
Subtract twice 9 + 18 + 27 +... 306

How many are equal to 18 / 24 in 1 / 4 15 / 20 9 / 12 3 / 4 25 / 100 75 / 100

The equivalent ones are:
15 / 20, 9 / 12, 3 / 4, 75 / 100, a total of four

The following table gives a "triangle number matrix" 12,34,5,67,8,9,10
It is known that there are n numbers in the nth row. Starting from the second row, the first number in each row is the last number of the previous row plus 1. Record the i-th row, and the j-th number is AI, J (I, J belong to n *)
1) Find a8,3
2) Try to write the expression of AI, J about I, J
1…… first line
3…… The second line
6…… The third line
10…… The fourth line
……
Have a professional one!

The first seven lines have 1 + 2 +. + 6 + 7 = 28 numbers, so a8,3 = 28 + 3 = 31
Ai,1=1+2+3+.i-2+i-1+1=i(i-1)/2+1
AI,J=i(i-1)/2+1+J-1=i(i-1)/2+J