第2个回答 2013-10-07
bn=(n+1)/(4an)=(n+1)/[4*2^(n-1)]=(n+1)/2^(n+1)
Tn=b1+b2+b3+……+bn
=(1+1)/2^(1+1)+(2+1)/2^(2+1)+(3+1)/2^(3+1)+……+(n+1)/2^(n+1)
=2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1) ①
2Tn=2*[2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1)]
=2/2^1+3/2^2+4/2^3+……+(n+1)/2^n ②
②-①
=2Tn-Tn
=【2/2^1+3/2^2+4/2^3+……+(n+1)/2^n 】-【2/2^2+3/2^3+4/2^4+……+(n+1)/2^(n+1)】
=2/2^1+1/2^2+1/2^3+……+1/2^n-(n+1)/2^(n+1)
=1+(1/2^2+1/2^3+……+1/2^n)-(n+1)/2^(n+1)
=1+1/4*(1-1/2^(n-1)/(1-1/2) -(n+1)/2^(n+1)
=1+1/8-1/2^(n+2)-(2n+2)/2^(n+2)
=9/8-(2n+3)/2^(n+2)本回答被网友采纳