设数列{an}满足a1=,an+1-an=3*2的2n-1 求数列{an通项公式 令bn=nan.求数列{bn}的前n项和
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![设数列{an}满足a1=,an+1-an=3*2的2n-1 求数列{an通项公式 令bn=nan.求数列{bn}的前n项和](/uploads/image/z/1963057-49-7.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D%2Can%2B1-an%3D3%2A2%E7%9A%842n-1+%E6%B1%82%E6%95%B0%E5%88%97%7Ban%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+%E4%BB%A4bn%3Dnan.%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C)
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设数列{an}满足a1=,an+1-an=3*2的2n-1 求数列{an通项公式 令bn=nan.求数列{bn}的前n项和
设数列{an}满足a1=,an+1-an=3*2的2n-1 求数列{an通项公式 令bn=nan.求数列{bn}的前n项和
设数列{an}满足a1=,an+1-an=3*2的2n-1 求数列{an通项公式 令bn=nan.求数列{bn}的前n项和
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
当n≥2时,a(n)=3·[2^1+2^3+...+2^(2n-3)]+2
=3·2[1-4^n]/(1-4) + 2
=2·4^n
当n=1时,a(n)=2
综上,a(n)=2·4^n .
b(n)=na(n)=2n·4^n
于是b(n)的前n项和为
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.
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