数学等比数列前n项和 紧急紧急!
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数学等比数列前n项和 紧急紧急!
数列an满足(1/2)*a1+(1/2^2)*a2+(1/2^3)*a3+...+(1/2^n)*an=5+2n,求an
数列an满足(1/2)*a1+(1/2^2)*a2+(1/2^3)*a3+...+(1/2^n)*an=5+2n,求an
![数学等比数列前n项和 紧急紧急!](/uploads/image/z/17453575-55-5.jpg?t=%E6%95%B0%E5%AD%A6%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8C+%E7%B4%A7%E6%80%A5%E7%B4%A7%E6%80%A5%21)
当n>1时
(1/2)(a1)+(1/2^2)(a2)+(1/2^3)(a3)+```+(1/2^n)(an)=2n+5 (1)
(1/2)(a1)+(1/2^2)(a2)+(1/2^3)(a3)+```+(1/2^(n-1))(a(n-1))=2(n-1)+5 (2)
(1)-(2)
则(1/2^n)(an)=2
an=2^(1+n)
当n=1的时候
1/2a1=2*1+5=7 a1=14
所以
a1=14
an=2^(1+n)(n>=2)
(1/2)(a1)+(1/2^2)(a2)+(1/2^3)(a3)+```+(1/2^n)(an)=2n+5 (1)
(1/2)(a1)+(1/2^2)(a2)+(1/2^3)(a3)+```+(1/2^(n-1))(a(n-1))=2(n-1)+5 (2)
(1)-(2)
则(1/2^n)(an)=2
an=2^(1+n)
当n=1的时候
1/2a1=2*1+5=7 a1=14
所以
a1=14
an=2^(1+n)(n>=2)