高中有点难度的数列设数列{an}的各项都是正数,且对任意n∈N*都有a1^3+a2^3+a3^3=(Sn)^2,记Sn为数列{an}的前n项和(1)求证:an^2=2Sn-an(2){an}的通项公式(3)若bn=3^n+(-1)^(n-1)*k*2^an(k为非零常数,n∈N*)问
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![高中有点难度的数列设数列{an}的各项都是正数,且对任意n∈N*都有a1^3+a2^3+a3^3=(Sn)^2,记Sn为数列{an}的前n项和(1)求证:an^2=2Sn-an(2){an}的通项公式(3)若bn=3^n+(-1)^(n-1)*k*2^an(k为非零常数,n∈N*)问](/uploads/image/z/1159635-3-5.jpg?t=%E9%AB%98%E4%B8%AD%E6%9C%89%E7%82%B9%E9%9A%BE%E5%BA%A6%E7%9A%84%E6%95%B0%E5%88%97%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E2%88%88N%2A%E9%83%BD%E6%9C%89a1%5E3%2Ba2%5E3%2Ba3%5E3%3D%28Sn%29%5E2%2C%E8%AE%B0Sn%E4%B8%BA%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%281%29%E6%B1%82%E8%AF%81%EF%BC%9Aan%5E2%3D2Sn-an%282%29%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%883%EF%BC%89%E8%8B%A5bn%3D3%5En%2B%28-1%29%5E%EF%BC%88n-1%EF%BC%89%2Ak%2A2%5Ean%28k%E4%B8%BA%E9%9D%9E%E9%9B%B6%E5%B8%B8%E6%95%B0%2Cn%E2%88%88N%2A%29%E9%97%AE)
高中有点难度的数列设数列{an}的各项都是正数,且对任意n∈N*都有a1^3+a2^3+a3^3=(Sn)^2,记Sn为数列{an}的前n项和(1)求证:an^2=2Sn-an(2){an}的通项公式(3)若bn=3^n+(-1)^(n-1)*k*2^an(k为非零常数,n∈N*)问
高中有点难度的数列
设数列{an}的各项都是正数,且对任意n∈N*都有a1^3+a2^3+a3^3=(Sn)^2,记Sn为数列{an}的前n项和
(1)求证:an^2=2Sn-an
(2){an}的通项公式
(3)若bn=3^n+(-1)^(n-1)*k*2^an(k为非零常数,n∈N*)问是否存在整数k,使得对任意n∈N*,都有bn+1大于bn
高中有点难度的数列设数列{an}的各项都是正数,且对任意n∈N*都有a1^3+a2^3+a3^3=(Sn)^2,记Sn为数列{an}的前n项和(1)求证:an^2=2Sn-an(2){an}的通项公式(3)若bn=3^n+(-1)^(n-1)*k*2^an(k为非零常数,n∈N*)问
1
A1^3+A2^3+A3^3+.+An^3=Sn^2
A1^3+A2^3+A3^3+.+A(n+1)^3=S(n+1)^2
两式相减,得
A(n+1)^3=(S(n+1)-Sn)(S(n+1)+Sn)
=A(n+1)(2S(n+1)-A(n+1))
所以
A(n+1)^2+A(n+1)=2S(n+1)
An^2+An=2Sn
2
两式相减,得
A(n+1)*(A(n+1)-1)=(An+1)*An
(A(n+1)+An)(A(n+1)-An-1)=0
因为An为正,所以有A(n+1)+An>0
A(n+1)=An+1
{An}为等差数列,公差为1
又A1^3=S1^2=A1^2
所以A1=1
所以得An通项为An=n
3
bn=3^n+(-1)^(n-1)*k*2^an
=3^n+(-1)^(n-1)*k*2^n
b(n+1)=3^(n+1)+(-1)^n*k*2^(n+1)
b(n+1)-bn
=2*3^n+(-1)^n*k*2^n*[2-1]
=2*3^n+(-1)^n*k*2^n
要使得b(n+1)-bn>0
2*3^n+(-1)^n*k*2^n>0
(-1)^n*k>-2*(3/2)^n
-2*(3/2)^n-3
即:
k >-3
-k>-3
所以:
-3