a1=1*(2+1)=3
設bn=nan
a1+2a2+3a3+…+nan=b1+b2+…+b(n-1)+bn=n(2n+1)
b1+b2+…+b(n-1)=(n-1)(2n-1)
兩式相減：
bn=n(2n+1)-(n-1)(2n-1)
=4n-1
an=bn/n=4-1/n;
設cn=nan/2^n,c1=a1/2=3/2
cn=(4n-1)/2^n
Tn=3/2+7/2^2+11/2^3+…+(4n-5)/2^(n-1)+(4n-1)/2^n
2Tn=3+7/2^1+11/2^2+…+(4n-5)/2^(n-2)+(4n-1)/2^(n-1)
兩式相減：
Tn=3+4/2^1+4/2^2+…+4/2^(n-1)-(4n-1)/2^n
=3+4[1/2^1+1/2^2+…+1/2^(n-1)]-(4n-1)/2^n
=3+4(1/2)[1/2^(n-1)-1]/(1/2-1)-(4n-1)/2^n
=3+4[1-1/2^(n-1)]-(4n-1)/2^n
=7-8/2^n-(4n-1)/2^n
=7-(4n+7)/2^n