数列an中有a1=1,an+1=1/3sn
数列an中有a1=1,an+1=1/3sn
求数列an的通项公式，求a2+a4+a6+a2n，【a后面的数字均为项数】
求数列an的通项公式，求a2+a4+a6+a2n，【a后面的数字均为项数】
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a(n+1)=(1/3)Sn
Sn = 3a(n+1)
an = Sn - S(n-1)
=3a(n+1) -3an
a(n+1) = (4/3)an
an = (4/3)^(n-1) .a1
= (4/3)^(n-1)
a2+a4+...+a(2n)
= (4/3)^1 + (4/3)^3+...+(4/3)^(2n-1)
=(4/3) [ (4/3)^(2n) -1] /(4/3-1]
=4[ (4/3)^(2n) -1]
Sn = 3a(n+1)
an = Sn - S(n-1)
=3a(n+1) -3an
a(n+1) = (4/3)an
an = (4/3)^(n-1) .a1
= (4/3)^(n-1)
a2+a4+...+a(2n)
= (4/3)^1 + (4/3)^3+...+(4/3)^(2n-1)
=(4/3) [ (4/3)^(2n) -1] /(4/3-1]
=4[ (4/3)^(2n) -1]
1年前
1
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