AlgebraDifficulty 5.6AIME, harderProve itUnited States
Problem: Let a1,a2,…,a1000 be real numbers such that a1⋅1+a2⋅2+a3⋅3+⋯+a1000⋅1000=0a1⋅12+a2⋅22+a3⋅32+⋯+a1000⋅10002=0a1⋅13+a2⋅23+a3⋅33+⋯+a1000⋅10003=0⋮a1⋅1999+a2⋅2999+a3⋅3999+⋯+a1000⋅1000999=0a1⋅11000+a2⋅21000+a3⋅31000+⋯+a1000⋅10001000=1. What is the value of a1 ?
Solution
Solution: The key idea is to multiply on weights which are polynomial coefficients. Specifically, let P(n)=n(n−2)(n−3)…(n−1000)=∑k=01000cknk, say. Note that c1=0 and c1000=1. Now take the kth equation and multiply it by ck, then add all equations together. From this we obtain a1P(1)+a2P(2)+⋯+a1000P(1000)=c1000=1. But P(1)=(−1)999⋅999! and P(2)=⋯=P(1000)=0. Therefore, a1=999!−1.
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