练习11:等差数列与等比数列 21.证明(1)由(3-m)S+2ma=m+3, nn 得(3-m)Sn+1+2man+1=m+3, 两式相减,得(3+m)a=2ma,m≠-3, n+1n a2m n+1 ∴=≠0(n≥1).∴{a}是等比数列. n m+3 a n (2)由(3-m)S+2ma=m+3,解出a=1,∴b=1. 1111 2m ? q=f(m)=,n∈N,且n≥2时, m+3 332b n?1 b=f(b)=·, nn-1 22 b+3 n?1 111 b+3b=3b,推出-=. bnn-1nn-1 bb3 nn?1 ?? 11 ∴是以1为首项、为公差的等差数列. ?? b3 ?n? 1n?1n+23 ∴=1+=.∴bn=. b33n+2 n 9/9chenpgb@126.com |
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