等差數(shù)列{an}:a1=1;an=n;
t1=b1=1;t2=2b1+b2=4;
b1=1;b2=2;
故等比數(shù)列{bn}:b1=1;bn=2^(n-1);
tn=nb1+an-1b2+.+a2bn-1+a1bn
=n+(n-1)*2+(n-2)*2^2+(n-3)*2^3+...+(n-(n-2))*2^(n-2)+(n-(n-1))*2^(n-1)
=n*(1+2+2^2+2^3+...+2^(n-2)+2^(n-1))-(1*2+2*2^2+3*2^3+...+(n-2)*2^(n-2)+(n-1)*2^(n-1))
=n*(2^n-1)-(n-2)*2^n-2
=2^(n+1)-n-2
減號(hào)后面的等差等比混合式子的計(jì)算如下：
令 x=1*2+2*2^2+3*2^3+...+(n-2)*2^(n-2)+(n-1)*2^(n-1) (1)
則2x= 1*2^2+2*2^3+...+(n-(n-3)*2^(n-2)+(n-(n-2))*2^(n-1)+(n-1)*2^n (2)
(2)-(1)得x=(n-1)*2^n-2^n+2=(n-2)*2^n+2