设数列{an}满足a1=2,an+1-an=3·2^(2n-1)令bn=n·an,求数列{bn}的前n项和Sn
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设数列{an}满足a1=2,an+1-an=3·2^(2n-1)令bn=n·an,求数列{bn}的前n项和Sn设数列{an}满足a1=2,an+1-an=3·2^(2n-1)令bn=n·an,求数列{
设数列{an}满足a1=2,an+1-an=3·2^(2n-1)令bn=n·an,求数列{bn}的前n项和Sn
设数列{an}满足a1=2,an+1-an=3·2^(2n-1)
令bn=n·an,求数列{bn}的前n项和Sn
设数列{an}满足a1=2,an+1-an=3·2^(2n-1)令bn=n·an,求数列{bn}的前n项和Sn
由递推式有
a2-a1=3*2
a3-a2=3*2*4
a4-a3=3*2*4^2
.
an-a(n-1)=3*2*4^(n-2)
累加得
an-a1=2*4^(n-1)-8
得an=2*4^(n-1)-6
于是bn=2n*4^(n-1)-6n
将其分为两部:2n*4^(n-1)与6n
后一部的sn=3n(n+1)
前一部的前n项和记为Tn,
则Tn=2+2*2*4+2*3*4^2+2*4*4^3+.+2*(n-1)*4^(n-2)+2*n*4^(n-1)
则4Tn=2*4+2*2*4^2+2*3*4^3+.+2*(N-1)*4^(n-1)+2*n*4^n
上式减下式得
-3Tn=2+2*4+2*4^2+.+2*4^(n-1)-2*n*4^n
即-3Tn=2(1+4+4^2.+4^(n-1))-2*n*4^n
即Tn=(2*n*4^n/3)-2*(4^n-1)/9
于是Sn=(2*n*4^n/3)-2*(4^n-1)/9-3n(n+1)
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