(1)設(shè){nan}數(shù)列的前n項和為Sn,則
Sn=a1+2a2+3a3+.+nan=n(2n+1)=2n^2+n
所以
S(n-1)=(n-1)[2(n-1)+1]
=2n^2-3n+1
所以
nan=Sn-S(n-1)
=4n-1
所以an=-1/n+4(n∈N+)
(2)由(1)得
nan=4n-1
所以
nan/(2^n)=4×n/(2^n)-1/(2^n)
所以
Tn=4[1/2+2/(2^2)+3/(2^3)+.+n/(2^n)]-[1/2+1/(2^2)+1/(2^3)+.+1/(2^n)]
令Fn=1/2+2/(2^2)+3/(2^3)+.+n/(2^n)
Gn=(1/2+1/(2^2)+1/(2^3)+.+1/(2^n)則
1/2Fn=1/(2^2)+2/(2^3)+3/(2^4)+.+(n-1)/(2^n)+n/[2^(n+1)]
Fn-1/2Fn=1/2+1/(2^2)+1/(2^3)+.+1/(2^n)-n/[2^(n+1)]
即
1/2Fn=-(2+n)/[2^(n+1)]+1
Fn=-(2+n)/(2^n)+2
Gn=1-1/(2^n)
所以
Tn=4Fn-Gn
=-(4n+7)/(2^n)+7
當n=1時,
T1=3/2=a1/2=4/2-1/2=3/2
所以
Tn=-(4n+7)/(2^n)+7(n∈N+)