己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三

己知公差大于零的等差数列an,a2＋a3＋a4＝9,且a2＋1,a3＋3,a4＋8为等比数列bn前三

1)假设公差为t则a3=a2+ta4=a2+2t==>3a2+3t=9==>a2+t=3==>a3=3又a2＋1,a3＋3,a4＋8为等比数列==》(3-t+1)(3+t+8)=36==>t=1==>an=n==>bn=3*2^(n-1)2)Sn=[(n+1)n]/2==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]又1/[(n+1)n]=1/n-1/(n+1)==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]=2[1-1/2+1/2-1/3+1/3-1/4.+n/1-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)======以下答案可供参考======供参考答案1:解设公差为d(1)a2=a1+d,a3=a1+2d,a4=a1+3da2＋a3＋a4＝a1+d+a1+2d+a1+3d=9 ,得3a1+6d=9,得a1+2d=a3=3,a2=3-d,a4=3+d(a2＋1):(a3＋3)=(a3＋3):(a4＋8) 得d=-8(不合题意舍去)或d=1所以an=n另：b1=a2+1=3,b2=a3+3=6,b3=a4+8=12,所以q=2,bn=3×2^(n-1)(2)Sn=[(n+1)n]/21/S1＋1/S2＋1/Sn=1+1/3+2/[(n+1)n]=

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