等比数列和等差数列题

显示全部楼层 · 2010-10-4 10:32:24

设数列an满足a1=2,an+1-an=3×2^(2n-1).
(1)求数列an的通项公式。（2）令bn=nan,求数列bn的前n项和Sn

千问 · 2010-10-4 10:32:24

(1)由a(n+1)-an=3×2^(2n-1)得an-a(n-1)=3×2^(2n-3)a(n-1)-a(n-2)=3×2^(2n-5)……a3-a2=3×2^3a2-a1=3×2^1上述(n-1)个式子累加得an-a1=3×2^1+3×2^3+……+3×2^(2n-5)+3×2^(2n-3)=3×2×[1-2^(n-1)]/(1-2）=6×[2^(n-1)-1]则an=a1+6×[2^(n-1)-1]=1+6×[2^(n-1)-1]=3*2^n-5(2)bn=nan=n(3*2^n-5)令cn=n*2^n则cn前n项和Mn=1*2^1+2*2^2+……+n*2^n （式1）同乘