如何重写带下标的求和符号并求解指定双重求和表达式

Let's break this down step by step—first clarifying the sigma notation structure, then solving your specific case with n=3 and given m values.

1. Re-expressing the Sigma Notation #

Your original expression is:
$$\sum_{j=0}^nj\sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j}$$

You’ve correctly split this into terms for each j from 0 to n. Here’s what each "something" corresponds to for every j:

• j=0: The inner sum is over empty index combinations (since we need 0 indices where (i_1 < ... < i_0)), which by convention has a value of 0. So this term is (0 * 0 = 0).
• j=1: The inner sum runs over all single-index combinations (each (i_1) from 1 to n). Each combination's sum is just (m_{i_1}), so the "something" here is (\sum_{i=1}^n m_i).
• j=2: The inner sum runs over all pairs of indices where (i_1 < i_2). For each pair, we add (m_{i_1} + m_{i_2}), so the "something" is (\sum_{1\le i_1 < i_2 \le n} (m_{i_1} + m_{i_2})).
• j=3: The inner sum runs over the only 3-index combination (since n=3): (i_1=1, i_2=2, i_3=3). The sum here is (m_1 + m_2 + m_3), so that's your "something" for j=3.

2. Calculating the Expression for n=3, (m_1=1), (m_2=2), (m_3=3) #

Let's compute each term individually:

• j=0 term: (0 * 0 = 0)
• j=1 term: (1 * (m_1 + m_2 + m_3) = 1*(1+2+3) = 6)
• j=2 term: First calculate the inner sum:
  $$(m_1+m_2) + (m_1+m_3) + (m_2+m_3) = (1+2)+(1+3)+(2+3) = 3+4+5 = 12$$
  Multiply by j=2: (2*12 = 24)
• j=3 term: (3*(m_1+m_2+m_3) = 3*(1+2+3) = 3*6 = 18)

Add all terms together: (0 + 6 + 24 + 18 = 48)

Quick Verification (Alternative Approach) #

Another way to check this result is to count how many times each (m_k) is included in the total sum:
For any (m_k), it appears in every j-index combination that includes k. The number of such combinations is (\binom{n-1}{j-1}) (choose j-1 more indices from the remaining n-1). Multiply that by j (the outer coefficient), and sum over j from 1 to n:
$$\sum_{j=1}^n j*\binom{n-1}{j-1} = (n+1)2^{n-2}$$
For n=3, this gives ((3+1)2^{1} = 8). So each (m_k) is counted 8 times:
$$8m_1 + 8m_2 +8m_3 =8(1+2+3)=8*6=48$$
Which matches our earlier calculation!

内容的提问来源于stack exchange，提问作者useranonis