数列(bn)是递增的等比数列,且b1+b3=5,b1b3=41.求数列(bn)的通项公式2.若an=log2 bn+3,求证数列(an)是等差数列3.若Cn=an*bn,求数列(Cn)的前n项和Sn
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/05 06:00:07
![数列(bn)是递增的等比数列,且b1+b3=5,b1b3=41.求数列(bn)的通项公式2.若an=log2 bn+3,求证数列(an)是等差数列3.若Cn=an*bn,求数列(Cn)的前n项和Sn](/uploads/image/z/3579400-64-0.jpg?t=%E6%95%B0%E5%88%97%EF%BC%88bn%EF%BC%89%E6%98%AF%E9%80%92%E5%A2%9E%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94b1%2Bb3%3D5%2Cb1b3%3D41.%E6%B1%82%E6%95%B0%E5%88%97%EF%BC%88bn%EF%BC%89%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2.%E8%8B%A5an%3Dlog2+bn%2B3%2C%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%EF%BC%88an%EF%BC%89%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%973.%E8%8B%A5Cn%3Dan%2Abn%2C%E6%B1%82%E6%95%B0%E5%88%97%28Cn%29%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
数列(bn)是递增的等比数列,且b1+b3=5,b1b3=41.求数列(bn)的通项公式2.若an=log2 bn+3,求证数列(an)是等差数列3.若Cn=an*bn,求数列(Cn)的前n项和Sn
数列(bn)是递增的等比数列,且b1+b3=5,b1b3=4
1.求数列(bn)的通项公式
2.若an=log2 bn+3,求证数列(an)是等差数列
3.若Cn=an*bn,求数列(Cn)的前n项和Sn
数列(bn)是递增的等比数列,且b1+b3=5,b1b3=41.求数列(bn)的通项公式2.若an=log2 bn+3,求证数列(an)是等差数列3.若Cn=an*bn,求数列(Cn)的前n项和Sn
设公比为q,b2=q*b1 ,b3=q^2*b1代入
{
b1+b3=5,b1*b3=4
}得b1=1,q=2;则{bn}的通项为bn=2^(n-1)
证明:an=log2 bn+3得an=n+2
an-a(n-1)=1即{an}是等差数列
cn=(n+2)*2^(n-1)
①Sn=3*1+4*2^1+.+(n+2)*2^(n-1)
②2sn= 3*2+4*2^2+..(n+1)*2^(n-1)+(n+2)*2^n
由②-①得:
Sn=-3+2+2^2+2^3+...+2^(n-1)+(n+2)*2^n
=-4+2^n+(n+2)*2^n=(n+3)*2^n-4
直接求得b1=1.b3=4然后容易得到通项bn=2^(n-1).a1=3,an-an-1=log2bn+1/bn=1.至于sn=a1b1+~anbn.2sn=a1b2+anbn+1作差得sn=(n+2)2^n-(2+~+2^n-1)-3=(n+3)2^n-5
(1)b1+b3=5,b1b3=4可得b1=1,b3=4或b1=4,b3=1
因为递增,所以b1=1,b3=4,得q=2或-2,因为递增,所以q=2
bn=2^(n-1)
(2)a(n+1)-an=log2bn+1-log2bn=log2q=1,所以{an}是等差数列
(3)a1=log2b1+3=0+3=3,得an=n+2
Cn=(n+2)*2^(n-1)...
全部展开
(1)b1+b3=5,b1b3=4可得b1=1,b3=4或b1=4,b3=1
因为递增,所以b1=1,b3=4,得q=2或-2,因为递增,所以q=2
bn=2^(n-1)
(2)a(n+1)-an=log2bn+1-log2bn=log2q=1,所以{an}是等差数列
(3)a1=log2b1+3=0+3=3,得an=n+2
Cn=(n+2)*2^(n-1)
等差乘等比,用错位相减
Sn=3*2^0+4*2^1+5*2^2+......+(n+2)*2^(n-1)
2Sn= 3*2^1+4*2^2+......+ (n+1)*2^(n-1)+(n+2)*2^n
两式相减
-Sn=3*2^0+{2^1+2^2+......+2^(n-1)}+(n+2)*2^n
化简后得
Sn={(n+1)*2^n}-1
收起
设公比为q,b2=q*b1 ,b3=q^2*b1代入
{
b1+b3=5,b1*b3=4
}得b1=1,q=2;则{bn}的通项为bn=2^(n-1)
证明:an=log2 bn+3得an=n+2
an-a(n-1)=1即{an}是等差数列
cn=(n+2)*2^(n-1)
①Sn=3*1+4*2^1+......+(n+2)*2...
全部展开
设公比为q,b2=q*b1 ,b3=q^2*b1代入
{
b1+b3=5,b1*b3=4
}得b1=1,q=2;则{bn}的通项为bn=2^(n-1)
证明:an=log2 bn+3得an=n+2
an-a(n-1)=1即{an}是等差数列
cn=(n+2)*2^(n-1)
①Sn=3*1+4*2^1+......+(n+2)*2^(n-1)
②2sn= 3*2+4*2^2+..(n+1)*2^(n-1)+(n+2)*2^n
由②-①得:
Sn=-3+2+2^2+2^3+...+2^(n-1)+(n+2)*2^n
=-4+2^n+(n+2)*2^n=(n+3)*2^n-4
收起