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AHSME 1968 Problem 17

    Let f(n)=x1+x2+⋯+xnnf(n)=\frac{x_1+x_2+\cdots +x_n}{n}, where nn is a positive integer. If xk=(−1)k,k=1,2,⋯ ,nx_k=(-1)^k, k=1,2,\cdots ,n, the set of possible values of f(n)f(n) is: (A) {0}(B) {1n}(C) {0,−1n}(D) {0,1n}(E) {1,1n}\text{(A) } \{0\} \text{(B) } \{\frac{1}{n}\} \text{(C) } \{0,-\frac{1}{n}\} \text{(D) } \{0,\frac{1}{n}\} \text{(E) } \{1,\frac{1}{n}\}

    Dataset text NuminaMath-1.5. Quoted with source.

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