设Tn=1/2^0+2/2+3/2^3+…+n/2^(n-1)(1)(1/2)*(1)得：(1/2)Tn=1/2+2/2^2+3/2^3+…+n/2^n(2)(1)-(2)得：(1/2)Tn=1+1/2+1/2^2+1/2^3+…+1/2^(n-1)-n/2^n=2-1/2^(n-1)-n/2^n这步不懂↑?

Question

设Tn=1/2^0+2/2+3/2^3+…+n/2^(n-1) (1)
(1/2)*(1)得：(1/2)Tn=1/2+2/2^2+3/2^3+…+n/2^n (2)
(1)-(2)得：
(1/2)Tn=1+1/2+1/2^2+1/2^3+…+1/2^(n-1)-n/2^n=2-1/2^(n-1)-n/2^n
这步不懂↑?

Answer 1

因为 an*a(n+2)=(n+1/2)(n+2+1/2).1 所以 1/an*1/a(n+2)=1/(n+1/2)* 1/(n+2+1/2)=[1/(n+1/2)-1/(n+2+1/2)]*1/2=tn.2 因为 tn-t(n-1)=[1/(n+1/2)-1/(n+2+1/2)]*1/2(n>1).3 所以t n=tn-t(n-1)+t(n-1)-t(n-2...