正项数列{an}中,n∈N+,且有2√Sn=an+1.求①an;②设{bn}=1/an•an+1.求{bn}前n项和Tn
人气:178 ℃ 时间:2020-03-26 19:43:46
解答
2√Sn=an+1变形为 4 Sn=(an+1)^2 ①4S(n-1)=(a(n-1)+1)^2 ② (下标我用括号括起来了)①-②得 4an=(an+1)^2 -(a(n-1)+1)^2 =(an)^2-(a(n-1))^2+2(an-a(n-1))整理可得 (an)^2-(a(n-1))^2-2(an+a(n...是{bn}=1/(an•an+1)你确定是{bn}=1/(an•an+1)不是{bn}=1/(an(an+1))或者是{bn}=1/(an*a(n+1))那我按{bn}=1/(an*a(n+1))来算,不是的话在完善当an是等比数列时bn=1/((-1)^(n-1)*(-1)^(n))=-1 bn是常数列所以Tn=-n当an是等差数列时bn=1/((2n-1)(2n+1))=1/2*(1/(2n-1)-1/(2n+1))则Tn=1/2*((1-1/3)+(1/3-1/5)+.........+1/(2n-1)-1/(2n+1))=1/2*(1-1/(2n+1))=n/(n+1)
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