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本文给出等差数列的两个判定方法,并举例说明其应用。 1.通项公式判定法:数列{a_n}为等差数列的充要条件是a_n=k_n+b.(k,b为常数) 证:若{a_n}是公差为d的等差数列,则a_n=a_1+(n-1)d=dn+(a_1-d),记d=k,a_1-d=b,∴a_n=kn+。若a_n=kn+b,(k,b为常数),则a_(n+1)-a_n=k(n+1)+b-(kn+l)=k, (n=1,2,…) 故{a_n}是等差数列。 2.前几项和判定法:数列{a_n}为等差数列的充要条件是S_n=an~2+bn,(a,b为常数) 证:若{a_n}是等差数列,则S_n=na_1+n(n-1)/2 d=(d/2)n~2+(2n_1-d)n/2
This article presents two methods of determining the arithmetic progression and illustrates its application. 1. General term formula judgment method: The necessary and sufficient condition for the sequence {a_n} to be an arithmetic progression is a_n=k_n+b. (k, b is a constant). Proof: If {a_n} is an arithmetic progression with tolerance d, then A_n=a_1+(n-1)d=dn+(a_1-d), note that d=k, a_1-d=b, and ∴a_n=kn+. If a_n=kn+b, (k, b is a constant), a_(n+1)-a_n=k(n+1)+b-(kn+l)=k, (n=1,2,... ) So {a_n} is an arithmetic progression. 2. The first few items and the determination method: The necessary and sufficient condition for the sequence {a_n} to be an arithmetic progression is S_n=an~2+bn, (a, b is a constant). Proof: If {a_n} is an arithmetic progression, S_n =na_1+n(n-1)/2 d=(d/2)n~2+(2n_1-d)n/2