a(n)-a(n-1)=3·2^(2n-3)
a(n-1)-a(n-2)=3·2^(2n-5)
...
a(2)-a(1)=3·2^1
a(1)=2
各式累加,有
當(dāng)n≥2時(shí),a(n)=3·[2^1+2^3+...+2^(2n-3)]+2
=3·2[1-4^n]/(1-4) + 2
=2·4^n
當(dāng)n=1時(shí),a(n)=2
綜上,a(n)=2·4^n .
b(n)=na(n)=2n·4^n
于是b(n)的前n項(xiàng)和為
S(n)=2·4+4·4²+6·4³+...+2n·4^n ...①
4S(n)=2·4²+4·4³+...+2(n-1)·4^n+2n·4^(n+1) ...②
①-②,有
-3S(n)=2·4+2(4²+4³+...+4^n)-2n·4^(n+1)
=8+2·[4²(1-4^n-1)/1-4]-2n·4^(n+1)
=8+32/3·[4^(n-1)-1]-2n·4^(n+1)
=8/3·4^n-8/3-8n·4^n
因此S(n)=(8n/3-8/9)·4^n-8/9.