已知数列{an}满足a1=1,a2=a(a≠0)an+2=p×(an+1)²/an(其他p为非零常数n∈N*)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/19 17:44:17
已知数列{an}满足a1=1,a2=a(a≠0)an+2=p×(an+1)²/an(其他p为非零常数n∈N*)判断数列{an+1/an}时不是等比数列(2)求an,(3)当a=1时,令bn=nan+2/an,Sn为数列bn的前n项和,求sn
a(n+2) = p[a(n+1)]^2/a(n),
a(n+2)/a(n+1) = p[a(n+1)/a(n)],
{a(n+1)/a(n)}是首项为a(2)/a(1)=a,公比为p的等比数列.
a(n+1)/a(n) = ap^(n-1),a(n)不为0.
a(n+1) = ap^(n-1)a(n),
a(n) = ap^(n-2)a(n-1),
a(n-1) = ap^(n-3)a(n-2),
...
a(3) = ap^(3-2)a(2)
a(2) = a*a(1),
a(n)a(n-1)...a(3)a(2) = a^(n-1)p^[1+2+...+(n-2)]a(n-1)a(n-2)...a(2)a(1),
a(n) = a^(n-1)p^[(n-1)(n-2)/2]a(1) = a^(n-1)p^[(n-1)(n-2)/2].
a=1时,a(n) = p^[(n-1)(n-2)/2].
a(n+2) = p^[(n+2-1)(n+2-2)/2] = p^[n(n+1)/2],
b(n) = na(n+2)/a(n) = np^[n(n+1)/2 - (n-1)(n-2)/2] = np^[(n^2+n - n^2 + 3n - 2)/2]
= np^(2n-1)
= np*p^(2n-2)
= np*(p^2)^(n-1).
s(n) = b(1)+b(2)+b(3)+...+b(n-1)+b(n)
= 1*p + 2*p*p^2 + 3*p*(p^2)^2 + ...+ (n-1)p*(p^2)^(n-2) + np*(p^2)^(n-1),
p=1时,s(n) = 1 + 2 + 3 + ...+ (n-1)+n = n(n+1)/2.
p不为1,且p不为0时,
p^2s(n) = 1*p*p^2 + 2*p*(p^2)^2 + ...+(n-1)p*(p^2)^(n-1) + np*(p^2)^n,
(1-p^2)s(n) = s(n) - p^2s(n) = 1*p + 1*p*p^2 + 1*p*(p^2)^2 + ...+ 1*p*(p^2)^(n-1) - np*(p^2)^n
= p[1+p^2 + ...+ (p^2)^(n-1)] - np(p^2)^n
= p[1 - (p^2)^n]/(1-p^2) - np(p^2)^n
= p[1 - p^(2n)]/(1-p^2 ) - np^(2n+1).
s(n) = p[1-p^(2n)]/(1-p^2)^2 - np^(2n+1)/(1-p^2).
综合,有,
(1) {a(n+1)/a(n)}是首项为a(2)/a(1)=a,公比为p的等比数列.
(2) a(n) = a^(n-1)p^[(n-1)(n-2)/2]a(1) = a^(n-1)p^[(n-1)(n-2)/2].
(3) p=1时,s(n) = n(n+1)/2.
p不为1,且不为0时,s(n) = p[1-p^(2n)]/(1-p^2)^2 - np^(2n+1)/(1-p^2).
a(n+2)/a(n+1) = p[a(n+1)/a(n)],
{a(n+1)/a(n)}是首项为a(2)/a(1)=a,公比为p的等比数列.
a(n+1)/a(n) = ap^(n-1),a(n)不为0.
a(n+1) = ap^(n-1)a(n),
a(n) = ap^(n-2)a(n-1),
a(n-1) = ap^(n-3)a(n-2),
...
a(3) = ap^(3-2)a(2)
a(2) = a*a(1),
a(n)a(n-1)...a(3)a(2) = a^(n-1)p^[1+2+...+(n-2)]a(n-1)a(n-2)...a(2)a(1),
a(n) = a^(n-1)p^[(n-1)(n-2)/2]a(1) = a^(n-1)p^[(n-1)(n-2)/2].
a=1时,a(n) = p^[(n-1)(n-2)/2].
a(n+2) = p^[(n+2-1)(n+2-2)/2] = p^[n(n+1)/2],
b(n) = na(n+2)/a(n) = np^[n(n+1)/2 - (n-1)(n-2)/2] = np^[(n^2+n - n^2 + 3n - 2)/2]
= np^(2n-1)
= np*p^(2n-2)
= np*(p^2)^(n-1).
s(n) = b(1)+b(2)+b(3)+...+b(n-1)+b(n)
= 1*p + 2*p*p^2 + 3*p*(p^2)^2 + ...+ (n-1)p*(p^2)^(n-2) + np*(p^2)^(n-1),
p=1时,s(n) = 1 + 2 + 3 + ...+ (n-1)+n = n(n+1)/2.
p不为1,且p不为0时,
p^2s(n) = 1*p*p^2 + 2*p*(p^2)^2 + ...+(n-1)p*(p^2)^(n-1) + np*(p^2)^n,
(1-p^2)s(n) = s(n) - p^2s(n) = 1*p + 1*p*p^2 + 1*p*(p^2)^2 + ...+ 1*p*(p^2)^(n-1) - np*(p^2)^n
= p[1+p^2 + ...+ (p^2)^(n-1)] - np(p^2)^n
= p[1 - (p^2)^n]/(1-p^2) - np(p^2)^n
= p[1 - p^(2n)]/(1-p^2 ) - np^(2n+1).
s(n) = p[1-p^(2n)]/(1-p^2)^2 - np^(2n+1)/(1-p^2).
综合,有,
(1) {a(n+1)/a(n)}是首项为a(2)/a(1)=a,公比为p的等比数列.
(2) a(n) = a^(n-1)p^[(n-1)(n-2)/2]a(1) = a^(n-1)p^[(n-1)(n-2)/2].
(3) p=1时,s(n) = n(n+1)/2.
p不为1,且不为0时,s(n) = p[1-p^(2n)]/(1-p^2)^2 - np^(2n+1)/(1-p^2).
已知数列{an}满足a1=4,an+1=an+p.3^n+1(n属于N+,P为常数),a1,a2+6,a3成等差数列.
已知数列{an}满足:a1=1,且an-a(n-1)=2n.求a2,a3,a4.求数列{an}通项an
已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
设数列[an}的前n项和为Sn,a1=a ,a2=p(p>0),Sn=n(an-a1)/2
已知数列an满足a1=2a,an=2a-a^2/an-1(n≥2)其中a是不为0的常数.求数列an的通项公式
如果数列an满足a{n+1}=pan+q(p,q为常数),则称an为"H数列".已知数列an的前n项和为Sn,若Sn=2
已知数列an满足条件a1=-2 a(n+1)=2an/(1-an) 则an=
关于数列极限的已知数列an满足a1=0 a2=1 an=(an-1+an-2)/2 求lim(n->无穷)an
已知数列{an}满足a1=1,a2=2,an+2=an+an+12,n∈N*.
数列an中,a1=1,a2=2数列bn满足an+1+(-1)n次an,a属于N* (1)若an等差数列...
已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1);