求n*Sₙ的公式：基于多层递推求和定义的数学推导问题

Let’s work through this problem step by step, starting from the basics to build up to the final expression. First, let’s make sure we’re clear on the custom notation given:

First, Recap the Definitions #

• 1*Sₙ is the standard sum of the first n natural numbers: 1 + 2 + ... + n
• For any k > 1, k*Sₙ means summing up (k-1)*S₁ all the way to (k-1)*Sₙ—so it’s a recursive "sum of sums" operation.

Important: The * here isn’t multiplication—it’s a custom operator for this recursive summation pattern.

Step 1: Calculate Base Cases to Spot a Pattern #

Let’s compute the first few k*Sₙ values using combinatorial notation (C(a,b) stands for "a choose b", which is a!/(b!(a-b)!)—this will make the pattern obvious):

1*Sₙ (Base Case) #

We all know the formula for the sum of the first n integers:

1*Sₙ = n(n+1)/2 = C(n+1, 2)

This is equivalent to choosing 2 elements from n+1 items—classic combinatorial interpretation.

2*Sₙ (Sum of 1*S₁ to 1*Sₙ) #

By definition, 2*Sₙ = 1*S₁ + 1*S₂ + ... + 1*Sₙ. Substitute the formula for 1*S_i:

2*Sₙ = sum_{i=1}^n C(i+1, 2)

There’s a key combinatorial identity here: the sum of C(r, k) from r=k to n equals C(n+1, k+1). Even when i=1 (where i+1=2), C(2,2)=1 fits, so we can apply this identity directly:

2*Sₙ = C(n+2, 3)

Let’s verify with n=2: 2*S₂ = 1*S₁ + 1*S₂ = 1 + 3 = 4, and C(4,3)=4—exact match.

3*Sₙ (Sum of 2*S₁ to 2*Sₙ) #

Following the rule, 3*Sₙ = sum_{i=1}^n 2*S_i = sum_{i=1}^n C(i+2,3). Apply the same identity:

3*Sₙ = C(n+3,4)

Check with n=2: 3*S₂ = 2*S₁ + 2*S₂ =1 +4=5, and C(5,4)=5—correct again.

Step 2: Generalize the Formula for k*Sₙ #

From these base cases, we can inductively prove that for any positive integer k:

k*Sₙ = C(n + k, k + 1)

Quick Inductive Proof #

• Base Case: We already confirmed k=1 gives C(n+1,2), which matches.
• Inductive Hypothesis: Assume for some k=m, m*Sₙ = C(n+m, m+1) is true.
• Inductive Step: For k=m+1, (m+1)*Sₙ = sum_{i=1}^n m*S_i = sum_{i=1}^n C(i+m, m+1). Using the combinatorial identity, this sum equals C(n+m+1, m+2)—which is exactly C(n + (m+1), (m+1)+1). So the formula holds for k=m+1.

Step 3: Derive n*Sₙ #

Now just substitute k=n into our general formula:

n*Sₙ = C(n + n, n + 1) = C(2n, n+1)

We can simplify this in a few ways using combination properties:

• Since C(a,b) = C(a, a-b), we can rewrite it as C(2n, n-1) (sometimes easier to work with).
• We can also express it in terms of the central binomial coefficient C(2n,n) (a common combinatorial value):

C(2n,n+1) = (n/(n+1)) * C(2n,n)

Or expand it fully with factorials:

n*Sₙ = (2n)! / [(n+1)! (n-1)!]

Example Check #

Let’s test n=3:

• 3*S₃ = 2*S₁ + 2*S₂ + 2*S₃ =1 +4 +10=15
• Using the formula: C(6,4)=15—perfect match.

内容的提问来源于stack exchange，提问作者Of course it's not me