摘要：21．解:(1)由an+1＝an+6an-1.an+1+2an＝3(an+2an-1) ∵a1＝5.a2＝5 ∴a2+2a1＝15故数列{an+1+2an}是以15为首项.3为公比的等比数列 ----4分得an+1+2an＝5?3n 由待定系数法可得(an+1-3n+1)＝-2(an-3n) 即an-3n＝2(-2)n-1 故an＝3n+2(-2)n-1＝3n-(-2)n ---8分(3)由3nbn＝n(3n-an)＝n[3n-3n+(-2)n]＝n(-2)n.∴bn＝n(-)n 令Sn＝|b1|+|b2|+-+|bn|＝+2()2+3()3+-+n()n Sn＝()2+2()3+-+n+n()n+1 ----10分得Sn＝+()2+()3+-+()n-n()n+1＝-n()n+1＝2[1-()n]-n()n+1 ∴ Sn＝6[1-()n]-3n()n+1＜6要使得|b1|+|b2|+-+|bn|＜m对于n∈N＊恒成立.只须m≥6 -12分

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