[Attachment] In order to facilitate editing and modification by those in need, the plain text document is provided as follows: Example 6. Given a natural number n, let X={1, 2, 3,..., Ar be r disjoint ternary elements of X subset, such that.

[Attachment] In order to facilitate editing and modification by those in need, the plain text document is provided as follows:

Example 6. Given a natural number n, let X={1, 2, 3,…,n}, A1, A2,…, Ar are r mutually disjoint ternary subsets of ≤r),

where S(A) represents the sum of all elements in A, and finds the maximum value of r.

analysis and solution: The meaning of division is |Ai|=3r.

and each A i are disjoint with each other, so S(Ai)=S(Ai)≥1+2+…+3r. According to conditions (1) and (2),

has S(Ai)≤n+(n–1)+…+(n–r+1),

so 1+2+…+3r≤n+(n–1) +...+(n–r+1), the solution is r≤(n–1).

notices that r∈N, so r≤[(n–1)].

If r=[(n–1)], we can construct r qualified ternary subsets. In order to satisfy the condition: S (Ai) ≤ n, the smallest possible element should be taken to belong to Ai, and try to construct Ai = {1, 2,..., 3r}. In order to satisfy the condition:

S(Ai)≠S(Aj), from the perspective of easy calculation, it can be imagined that {S(Ai)} is an arithmetic sequence with a tolerance of 1. In this way, we only need to divide 1, 2,...,2r into r sets first, so that their sums are equal. This is easy to do using "size matching". Finally, 2r+1, 2r+2, 2r+3,..., 3r are allocated to each set, and one element is allocated to each set. For example, let Ai={i, 2r+i, 2r–i+1} (i=1, 2,…,r), then the sum of each set is not equal, and the maximum sum is 5r+1≤(n– 1)+1=n.

Therefore, the maximum value of r is [(n–1)].

Example 7. Verification: For any 0ab, there are positive integers k and n1, n2,...,nk, such that a++...+b. Analysis and solution of

: First, in order to make ++...+ easy to estimate, notice that the series is divergent, and think of finding continuous natural numbers n, n+1, n+2,..., n+k, so that inequality (* ) was established. Due to the divergence, we can first find k so that a is established (that is, the left end of the inequality is established). In order to make the right side of the inequality also true, it should be as small as possible, so k is the smallest k that satisfies a, that is, a, and ≤ a. In this way, a=+≤a+.

we expect a+b, a sufficient condition for is a+b, which is n.

but a≤, so n.

In summary, it can be seen that taking nmax{, }, the inequality is proved.

Note: Taking a=33–π–1959 and b=33–π–1992, we get 33 alternative questions: There are natural numbers k and mutually different natural numbers: n1, n2,…,nk, so that π–199233–( ++…+)π–1959. It can be seen from this question that the constants 33, π–1992, and π–1959 in the question are all "deceptive".