已知数列满足{an}满足a1=2,an+1=an-{n(n+1)分之一}
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已知数列满足{an}满足a1=2,an+1=an-{n(n+1)分之一}
1求数列{an}的通向公式
2设bn=n 乘an2的n次方.求数列{bn}的前项和Sn
是an乘(2的n次方)
1求数列{an}的通向公式
2设bn=n 乘an2的n次方.求数列{bn}的前项和Sn
是an乘(2的n次方)
1)累加法
a1=2
a2-a1=1/(1*2)
a3-a2=1/(2*3)
a4-a3=1/(3*4)
.
an-a(n-1)=1/[(n-1)n]
相加得
an=2+(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+[1/(n-1)-1/n]
=2+(1-1/n)=3-1/n=(3n-1)/n.
2)bn=n*an*2^n=(3n-1)*2^n,(错位相减)
Sn=2*2^1+5*2^2+8*2^3+.+(3n-1)*2^n
2Sn= 2*2^2+5*2^3+.+(3n-4)*2^n+(3n-1)*2^(n+1)
相减得
Sn=-4-3(2^2+2^3+...+2^n)+(3n-1)*2^(n+1)
=-4-3*[2^(n+1)-4]+(3n-1)*2^(n+1)
=(3n-4)*2^(n+1)+8.
a1=2
a2-a1=1/(1*2)
a3-a2=1/(2*3)
a4-a3=1/(3*4)
.
an-a(n-1)=1/[(n-1)n]
相加得
an=2+(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+[1/(n-1)-1/n]
=2+(1-1/n)=3-1/n=(3n-1)/n.
2)bn=n*an*2^n=(3n-1)*2^n,(错位相减)
Sn=2*2^1+5*2^2+8*2^3+.+(3n-1)*2^n
2Sn= 2*2^2+5*2^3+.+(3n-4)*2^n+(3n-1)*2^(n+1)
相减得
Sn=-4-3(2^2+2^3+...+2^n)+(3n-1)*2^(n+1)
=-4-3*[2^(n+1)-4]+(3n-1)*2^(n+1)
=(3n-4)*2^(n+1)+8.
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