已知数列an满足:a1=2,a(n+1)=3an+3^(n+1)-2^n(1)设bn=(an-2^n)/3^n,证明bn为等差数列,并求an通项公式(2)求an前n项和Sn
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![已知数列an满足:a1=2,a(n+1)=3an+3^(n+1)-2^n(1)设bn=(an-2^n)/3^n,证明bn为等差数列,并求an通项公式(2)求an前n项和Sn](/uploads/image/z/12461842-10-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3%3Aa1%3D2%2Ca%28n%2B1%29%3D3an%2B3%5E%28n%2B1%29-2%5En%EF%BC%881%EF%BC%89%E8%AE%BEbn%3D%28an-2%5En%29%2F3%5En%2C%E8%AF%81%E6%98%8Ebn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E6%B1%82an%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
已知数列an满足:a1=2,a(n+1)=3an+3^(n+1)-2^n(1)设bn=(an-2^n)/3^n,证明bn为等差数列,并求an通项公式(2)求an前n项和Sn
已知数列an满足:a1=2,a(n+1)=3an+3^(n+1)-2^n
(1)设bn=(an-2^n)/3^n,证明bn为等差数列,并求an通项公式(2)求an前n项和Sn
已知数列an满足:a1=2,a(n+1)=3an+3^(n+1)-2^n(1)设bn=(an-2^n)/3^n,证明bn为等差数列,并求an通项公式(2)求an前n项和Sn
(1)
a(n+1)=3an+3^(n+1)-2ⁿ
a(n+1)-2^(n+1)=3an+3^(n+1)-2ⁿ-2^(n+1)=3an+3^(n+1)-3×2ⁿ=3(an -2ⁿ) +3^(n+1)
[a(n+1)-2^(n+1)]/3^(n+1)=(an -2ⁿ)/3ⁿ +1
[a(n+1)-2^(n+1)]/3^(n+1)-(an-2ⁿ)/3ⁿ=1,为定值
(a1-2)/3=(2-2)/3=0
数列{(an-2ⁿ)/3ⁿ}是以0为首项,1为公差的等差数列
bn=(an-2ⁿ)/3ⁿ,数列{bn}是以0为首项,1为公差的等差数列
(an-2ⁿ)/3ⁿ=0+1×(n-1)=n-1
an=(n-1)×3ⁿ+2ⁿ
数列{an}的通项公式为an=(n-1)×3ⁿ+2ⁿ
(2)
Sn=a1+a2+...+an=0×3+1×3²+2×3³+...+(n-1)×3ⁿ +(2+2²+...+2ⁿ)
令An=0×3+1×3²+2×3³+...+(n-1)×3ⁿ
则3An=0×3²+1×3³+...+(n-2)×3ⁿ+(n-1)×3^(n+1)
An-3An=-2An=3²+3³+...+3ⁿ-(n-1)×3^(n+1)
=9×[3^(n-1)-1]/(3-1) -(n-1)×3^(n+1)
=[(3-2n)×3^(n+1) -9]/2
An=[(2n-3)×3^(n+1)+9]/4
Sn=An+(2+2²+...+2ⁿ)
=[(2n-3)×3^(n+1)+9]/4+ 2×(2ⁿ-1)/(2-1)
=[(2n-3)×3^(n+1)+2^(n+3) +1]/4