已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81(1)求数列的前三项a1,a2,a3;(2)数列{(an+p)/2^n}为等差数列,求实数p的值;(3)求数列{an}的前n项和S
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![已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81(1)求数列的前三项a1,a2,a3;(2)数列{(an+p)/2^n}为等差数列,求实数p的值;(3)求数列{an}的前n项和S](/uploads/image/z/9320063-23-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3an%3D2a%28n-1%29%2B2%5En-1%28n%E2%88%88N%2B%2C%E4%B8%94n%3E%3D2%29%2Ca4%3D81%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E5%89%8D%E4%B8%89%E9%A1%B9a1%2Ca2%2Ca3%EF%BC%9B%EF%BC%882%EF%BC%89%E6%95%B0%E5%88%97%7B%28an%2Bp%29%2F2%5En%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%E5%AE%9E%E6%95%B0p%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%883%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CS)
已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81(1)求数列的前三项a1,a2,a3;(2)数列{(an+p)/2^n}为等差数列,求实数p的值;(3)求数列{an}的前n项和S
已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81
(1)求数列的前三项a1,a2,a3;
(2)数列{(an+p)/2^n}为等差数列,求实数p的值;
(3)求数列{an}的前n项和S
已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81(1)求数列的前三项a1,a2,a3;(2)数列{(an+p)/2^n}为等差数列,求实数p的值;(3)求数列{an}的前n项和S
an=2a(n-1)+2^n-1
an/2^n=a(n-1)/2^(n-1) +1-1/2^n
an/2^n-a(n-1)/2^(n-1)=1-1/2^n
a2/2^2-a1/2=1-1/2^2
a3/2^3-a2/2^2=1-1/2^3
...
an/2^n-a(n-1)/2^(n-1)=1-1/2^n
左右两边分别相加:
an/2^n-a1/2=(n-1)-[1/2^2+..+1/2^n]
an/2^n=a1/2+(n-1)-1/2^2*[1-1/2^(n-1)]/(1-1/2)
=a1/2+(n-1)-(1/2-1/2^n)
=a1/2+n-3/2+1/2^n
a1=5 (a4求a3-a2-a1,
an=(n+1)*2^n+1
Sn=a1+a2+..+an=n+2^1+2^2+..+2^n+1*2^1+2*2^2+..+n*2^n
=n+2*(1-2^n)/(1-2)+1*2^1+2*2^2+..+n*2^n
bn=1*2^1+2*2^2+..+n*2^n
2bn=1*2^2+.+(n-1)*2^n+n*2^(n+1)
相减:
-bn=2+2^2+.+2^n-n*2^(n+1)
=2*(1-2^n)/(1-2)-n*2^(n+1)
bn=-2*(1-2^n)/(1-2)+n*2^(n+1)
Sn=n+n*2^(n+1)
(1)令n=4,a4=2a3+16-1
a3=33 类似可以退出:a2=13 a1=5
(2)用待定系数法:(an+p)/2^n=(a(n-1)+p)/2^(n-1)+k(k为常数)
推导得:an=2a(n-1)+p+k*2^n
比较系数得:p=-1,k=1
(3)不妨设bn=(an-1)/2^n
由(2)可得:bn=b(n-1)+1=b1...
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(1)令n=4,a4=2a3+16-1
a3=33 类似可以退出:a2=13 a1=5
(2)用待定系数法:(an+p)/2^n=(a(n-1)+p)/2^(n-1)+k(k为常数)
推导得:an=2a(n-1)+p+k*2^n
比较系数得:p=-1,k=1
(3)不妨设bn=(an-1)/2^n
由(2)可得:bn=b(n-1)+1=b1+(n-1)
b1=(a1-1)/2=2
所以bn=n+1
故n+1=(an-1)/2^n
an=(n+1)*2^n+1
Sn=2*2^1+3*2^2+4*2^3+···+(n+1)*2^n+n
(下面用到错位相减)
2Sn= 2*2^2+3*2^3+···+ n*2^n+(n+1)*2^(n+1)+2n
两式相减:
Sn=-(4+2^2+2^3+···+2^n)+(n+1)*2^(n+1)+n
整理得:
Sn=n*(2(n+1)+1)
(n>=1)
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