According to the natural number from small to large as the standard order, find the following permutation of reverse order number 25 （3n-1）3 6… （3(n-1））（3n)

According to the natural number from small to large as the standard order, find the following permutation of reverse order number 25 （3n-1）3 6… （3(n-1））（3n)

（3n-25）+（3n-26）+.+1=（3n-24）*（3n-25）/2

According to the standard order of natural number from small to large, find 1, 3 (2n-1) 2 4… 2n

Because odd numbers and even numbers are arranged from small to large, an odd number and an even number can form a reverse order pair,
There are 1 pair of 3, 2 pairs of 5 If 2N-1 is contained, there are n-1 pairs,
So inverse ordinal = 1 + 2 + 3 + +(n-1)= n(n-1)/2

Let n be a natural number, [x] denote the largest integer not exceeding X
x+2【x】+3【3】+【4】+.+n【x】=n（n+1）×n(n+1）

Gauss equation, eh. But there's something wrong with your question. Is the first [x] and the third and fourth terms 3 [x] and 4 [x]