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依據是:等差數列中等距的兩項乘積的倒數數列均可以用裂項相消法求和如:1/n*(n+1)1/(2n-1)*(2n+1)1/an*a(n+1)1/an*a(n+k)如何裂開1/an*a(n+k)方法:逆求法:將數列的通項裂開1/an-1/a(n+k)=(a(n+k)-an)/[an*a(n+k)]...
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