已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^n+1-n-2已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^(n+1)-n-21、若数列{an}是首项和公差都是1
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![已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^n+1-n-2已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^(n+1)-n-21、若数列{an}是首项和公差都是1](/uploads/image/z/5931319-31-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%7Bbn%7D%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0N%2C%E9%83%BD%E6%9C%89%EF%BC%9Aa1bn%2Ba2bn-1%2Ba3bn-2%2B%E2%80%A6%E2%80%A6%2Ban-1b2%2Banb1%3D2%5En%2B1-n-2%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%7Bbn%7D%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0N%2C%E9%83%BD%E6%9C%89%EF%BC%9Aa1bn%2Ba2bn-1%2Ba3bn-2%2B%E2%80%A6%E2%80%A6%2Ban-1b2%2Banb1%3D2%5E%28n%2B1%29-n-21%E3%80%81%E8%8B%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E9%A6%96%E9%A1%B9%E5%92%8C%E5%85%AC%E5%B7%AE%E9%83%BD%E6%98%AF1)
已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^n+1-n-2已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^(n+1)-n-21、若数列{an}是首项和公差都是1
已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^n+1-n-2
已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^(n+1)-n-2
1、若数列{an}是首项和公差都是1的等差数列,求证数列{bn}是等比数列
2、若数列{bn}是等比数列,证明当q=2时,数列{an}是等差数列
已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^n+1-n-2已知数列{an}{bn},对任意正整数N,都有:a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1=2^(n+1)-n-21、若数列{an}是首项和公差都是1
1.an=n,
设 Sn=a1bn+a2bn-1+a3bn-2+……+an-1b2+anb1,则:
S_(n)=b_(n)+2*b_(n-1)+3*b_(n-2)+...+(n-1)*b_(2)+n*b_(1)=2^(n+1)-n-2------------------1
S_(n-1)=b_(n-1)+2*b_(n-2)+3*b_(n-3)+...+(n-2)*b(2)+(n-1)*b1=2^(n)-n-1------------------2
用式子1减去式子2,得:
Bn=b_(n)+b_(n-1)+...+b_(1)=2^(n)-1
Bn为 数列{bn}的前n项和.所以数列{bn}的通项为:
b_(n)=B_(n)-B_(n-1)=2^(n)-2^(n-1)=2^(n-1)
所以,数列{bn}为首项为1,公比为2的等比数列.
2.设数列{bn}为首项为\x1db,公比为q的等比数列.则:
S_(n)=b*[a_(1)*q^(n)+a_(2)*q^(n-1)+...+a_(n)]=2^(n+1)-n-2----------------------------1
S_(n-1)=b*[a_(1)*q^(n-1)+a_(2)*q^(n-2)+...+a_(n-1)]=2^(n)-n-1
q*S_(n-1)=b*[a_(1)*q^(n)+a_(2)*q^(n-1)+...+a_(n-1)*q]=q*[2^(n)-n-1]----------------------2
用式子1减去式子2,得:
a_(n)=[(2-q)/b]*2^(n)+[(q-1)/b]*(n-1)+(2q-3)/b
若要使得数列{an}为等差数列,则通项应有形式:
a_(n)=a_(1)+(n-1)*d
所以 q=2的时候,数列{an}为等差数列.首项为 1/b,公差为 1/b.
hf
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