已知数列{an}满足a1=1,a(n+1)=2an+1(n∈N)

显示全部楼层 · 2011-10-1 00:08:08

a(n+1)=2an+1a(n+1)+1=2[an+1][a(n+1)+1]/[an+1]=2等比公比为2首项2an+1=2*2(n-1)=2^nan=2^n-1(2)4^(b1+2b2+3b3+……+nbn－n) =(2^n)^n=2^n*n 2(b1+2b2+3b3+……+nbn－n)=n^2 Bn=b1+2b2+3b3+……+nbn=1/2×n^2+nBn-Bn-1=nbn=n+1/2bn=1/2n+1（3）若cn=2^n/(ana(n+1))，求数列{cn}的前n项和Sncn=2^n/(ana(n+1))=2^n/(2^n-1)[2^(n+1)-1]=1/(2^n-1)-1/[2^(n+1)-...

千问 · 2011-10-1 00:08:08

(1)a(n+1)+k=2×(an+k) ，展开与已知式比较，由对应项相等得k=1所以an+1为以2为首项、2为公比的等比数列，所以an=2^n－1(2)原式左边=4^(b1+2b2+3b3+……+nbn－n) 右边=(2^n)^n=2^n^2则 2(b1+2b2+3b3+……+nbn－n)=n^2则记Bn=b1+2b2+3b...

千问 · 2011-10-1 00:08:08

两边同时加上1/2-1=1 即a(n+1)+1=2an+1+1 即a(n+1)+1=2（an+1）即[a(n+1)+1]/an+1=2则数列｛an+1｝是等比数列，公比是2，首项是2.所以an+1=2*2^(n-1) 所以an=2*2^(n-1)+1...