高手请帮我解答数列练习题设数列{an}a1=1,前N项的和Sn满足3tSn-(2t+3)Sn-1=3t(t〉0,n=2,3,4…)求证:(1) 数列{an}是等比数列(2) 设{an}公比为f(1/bn-1),作数列{bn}使b1=1,bn=f(1/bn-1)(n=2,3,4…)
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![高手请帮我解答数列练习题设数列{an}a1=1,前N项的和Sn满足3tSn-(2t+3)Sn-1=3t(t〉0,n=2,3,4…)求证:(1) 数列{an}是等比数列(2) 设{an}公比为f(1/bn-1),作数列{bn}使b1=1,bn=f(1/bn-1)(n=2,3,4…)](/uploads/image/z/1270332-36-2.jpg?t=%E9%AB%98%E6%89%8B%E8%AF%B7%E5%B8%AE%E6%88%91%E8%A7%A3%E7%AD%94%E6%95%B0%E5%88%97%E7%BB%83%E4%B9%A0%E9%A2%98%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9Da1%3D1%2C%E5%89%8DN%E9%A1%B9%E7%9A%84%E5%92%8CSn%E6%BB%A1%E8%B6%B33tSn-%282t%2B3%29Sn-1%3D3t%28t%E3%80%890%2Cn%3D2%2C3%2C4%E2%80%A6%29%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89+%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89+%E8%AE%BE%EF%BD%9Ban%EF%BD%9D%E5%85%AC%E6%AF%94%E4%B8%BAf%281%2Fbn-1%29%2C%E4%BD%9C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E4%BD%BFb1%3D1%2Cbn%3Df%281%2Fbn-1%29%28n%3D2%2C3%2C4%E2%80%A6%29)
高手请帮我解答数列练习题设数列{an}a1=1,前N项的和Sn满足3tSn-(2t+3)Sn-1=3t(t〉0,n=2,3,4…)求证:(1) 数列{an}是等比数列(2) 设{an}公比为f(1/bn-1),作数列{bn}使b1=1,bn=f(1/bn-1)(n=2,3,4…)
高手请帮我解答数列练习题
设数列{an}a1=1,前N项的和Sn满足3tSn-(2t+3)Sn-1=3t(t〉0,n=2,3,4…)求证:
(1) 数列{an}是等比数列
(2) 设{an}公比为f(1/bn-1),作数列{bn}使b1=1,bn=f(1/bn-1)(n=2,3,4…)
(3) 求和b1b2-b2b3+…+b2n-1b2n-b2nb2n+1
高手请帮我解答数列练习题设数列{an}a1=1,前N项的和Sn满足3tSn-(2t+3)Sn-1=3t(t〉0,n=2,3,4…)求证:(1) 数列{an}是等比数列(2) 设{an}公比为f(1/bn-1),作数列{bn}使b1=1,bn=f(1/bn-1)(n=2,3,4…)
这是个高考题啊,分太少,无视
(1)∵3t*Sn-(2t+3)S(n-1)=3t,3t*[S(n-1)+an]-(2t+3)S(n-1)=3t,
∴(t-3)S(n-1)+3tan=3t…①,(t-3)Sn+3ta(n+1)=3t…②,
②-①得,(t-3)[Sn-S(n-1)]+3t[a(n+1)-an]=0=
(t-3)(an)+3t[a(n+1)-an]=0,
∴a(n+1)/an...
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(1)∵3t*Sn-(2t+3)S(n-1)=3t,3t*[S(n-1)+an]-(2t+3)S(n-1)=3t,
∴(t-3)S(n-1)+3tan=3t…①,(t-3)Sn+3ta(n+1)=3t…②,
②-①得,(t-3)[Sn-S(n-1)]+3t[a(n+1)-an]=0=
(t-3)(an)+3t[a(n+1)-an]=0,
∴a(n+1)/an=(2t+3)/3t,
∴{an}是等比数列。
(2)∵bn=3b(n-1)/2b(n-2)+3
∴两边求倒:1/bn=2/3+1/b(n-1)
∴{1/bn}为公差2/3的等差数列
∴bn=(2n+1)/3
(3)因为bn公差为2/3 ,所以,b(k+2)-b(k)=4/3,{b2n}是以b2=5/3为首项,4/3为公差的等差数列
原式=b1b2-b2b3+b3b4-b4b5-...+b2n-1b2n-b2nb2n+1
=b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1 - b2n+1 )
=(-4/3)(b2+b4+…+b2n)
=(-4/3){5n/3+[n*(n-1)/2]*4/3}
=-(8n^2+12n)/9
收起