设 S = {1,2,3,..., n}, 令 f(S) 表示所有 S 的非空子集和的平方和, 即:...
Créé le : 22 janvier 2025
Créé le : 22 janvier 2025
设 S = {1,2,3,..., n}, 令 f(S) 表示所有 S 的非空子集和的平方和, 即:
求当 n = 2023 时, f(S) mod (10^9 + 7) 的值。
Summing squares of subset sums
OK, let's think this through: we're calculating the sum of squares of sums for all non-empty subsets of a set , and then take it modulo .
Breaking it down
I’m working through the identity for by expanding and separating sums, examining each in , and considering all non-empty subsets .
Confirming subset contributions
I’m verifying that each element in appears in exactly non-empty subsets, ensuring the sum of across all accurately reflects these contributions.
Summing pairs
I’m gathering info on the sum of products and over all non-empty subsets , considering how each distinct pair appears in subsets.
Summing distinct pairs
I’m adding up the products of all distinct pairs and from , ensuring each pair is counted only once in the sum.
Crafting the formula
I’m working on simplifying by unifying the sums of squares for each in with the sum of products for distinct pairs , leading to a concise expression.
Breaking down the approach
I’m simplifying the expression by noting that the sum of squares plus the sum of products for distinct pairs is half the sum of the squares plus the square of the sum of .
Combining and simplifying
I’m amalgamating terms, isolating common factors, and streamlining the algebraic expression to deduce a simpler closed-form solution for .
Finalizing the expression
I’m adding the final touches to a simplified version of , factoring into , and ensuring all terms are neatly combined for a clearer form.
Finalizing the steps
I’m mapping out the calculation for modulo by carefully breaking down the terms, ensuring each part is accurately computed and combined to find the final result.
Isolating powers
I’m piecing together the factorization of large numbers to simplify modulo , isolating common factors and simplifying the division by 12.
Breaking down calculations
I’m meticulously assembling the factorization of large numbers to streamline modulo , isolating common factors, and making the division by 12 easier.
Calculating powers
I’m multiplying prime factors of large numbers to simplify modulo , accumulating powers of 2, 3, 5, 7, 17, 11, 23, and 607 to streamline the division by 12.
Pulling together
OK, let me see. I’m combining divided by 12 with to form , then aligning these terms for modulo .
Ensuring precision
I’m moving through the steps to compute modulo , ensuring each term is accurately factored and multiplied, while maintaining the integrity of the factors.
Breaking down the exponentiation
OK, let me see. I’m thinking about a method to compute modulo by breaking it into smaller parts and using modular exponentiation techniques.