Sum of the products of the subsets

Algebra Level 4

Consider the set { 1 , 2 , 3 , . . . 2016 , 2017 } \{1, 2, 3, ... 2016, 2017\} . It contains all the positive integers from 1 to 2017. The total sum of the products of the elements in each non-empty subset of this set is S S . Suppose f ( x ) f(x) is a function such that for positive integers x x , f ( x ! ) = x f(x!) = x . Find f ( S + 1 ) f(S+1) .


The answer is 2018.

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1 solution

Chew-Seong Cheong
May 15, 2017

Consider the following product:

k = 1 n ( x + a k ) = x n + x n 1 k = 1 n a k + x n 2 j , k = 1 , j k n a j a k + x n 3 i , j , k = 1 , i j k n a i a j a k + + k = 1 n a k = x n + c n 1 x n 1 + c n 2 x n 2 + c n 3 x n 3 + + c 2 x 2 + c 1 x + c 0 \begin{aligned} \prod_{k=1}^n \left(x+a_k\right) & = x^n + x^{n-1} \sum_{k=1}^n a_k + x^{n-2} \sum_{j,k=1, j \ne k}^n a_ja_k + x^{n-3} \sum_{i,j,k=1, i \ne j \ne k}^n a_ia_ja_k + \cdots + \prod_{k=1}^n a_k \\ & = x^n + c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + c_{n-3}x^{n-3} + \cdots + c_2 x^2 + c_1 x + c_0 \end{aligned}

We note that k = 0 n 1 c k = S a \displaystyle \sum_{k=0}^{n-1} c_k = S_a is the sum of products of each non-empty subset of the set { a 1 , a 2 , a 3 , . . . a n } \{a_1, a_2, a_3, ... a_n\} . Replacing x = 1 x=1 and a k = k a_k = k , we have:

k = 1 n ( 1 + k ) = 1 + S S + 1 = ( n + 1 ) ! For n = 2017 S + 1 = 2018 ! f ( S + 1 ) = 2018 \begin{aligned} \prod_{k=1}^n \left(1+k\right) & = 1 + S \\ \implies S+1 & = (n+1)! & \small \color{#3D99F6} \text{For }n = 2017 \\ S + 1 & = 2018! \\ \implies f(S+1) & = \boxed{2018} \end{aligned}

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