已知α,β是方程x²+2002x+1=0的两根,则(1+2003α+α²)(1+2003β+β²)=_________________再加一体 观察:0、3、8、15、24……则它的第2003个数是__________________
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![已知α,β是方程x²+2002x+1=0的两根,则(1+2003α+α²)(1+2003β+β²)=_________________再加一体 观察:0、3、8、15、24……则它的第2003个数是__________________](/uploads/image/z/648624-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%2C%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8Bx%26%23178%3B%2B2002x%2B1%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C%E5%88%99%281%2B2003%CE%B1%2B%CE%B1%26%23178%3B%EF%BC%89%EF%BC%881%2B2003%CE%B2%2B%CE%B2%26%23178%3B%EF%BC%89%3D_________________%E5%86%8D%E5%8A%A0%E4%B8%80%E4%BD%93+%E8%A7%82%E5%AF%9F%EF%BC%9A0%E3%80%813%E3%80%818%E3%80%8115%E3%80%8124%E2%80%A6%E2%80%A6%E5%88%99%E5%AE%83%E7%9A%84%E7%AC%AC2003%E4%B8%AA%E6%95%B0%E6%98%AF__________________)
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