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设公差不为0的等差数列{an}与等比数列{bn}q为正满足a1=b1=1,a3=b3,a7=b5, 1 求等比数列q的值 2 记Mn=a1+a2+a3+------an Nn=b1+b2+b3+-----bn试比较M5 N5 的大小 3若a1=1设数列Cn=a﹙2n+1﹚×b﹙2n+1﹚求数列﹛Cn﹜的前n项和 谢谢了

Answer 1

1、设an=a1+(n-1)d=1+(n-1)d,bn=b1*q^(n-1)=q^(n-1),则a3=1+2d,a7=1+6d,b3=q^2,b5=q^4
由a3=b3,a7=b5,得1+2d=q^2……①,1+6d=q^4……②
由①式得，d=（q^2-1）/2,代入②式，得q^4-3q^2+2=0,解得，q^2=1或q^=2,因为an公差不为0，既a1≠a3,b1≠b3,所以q^2=2,q=√2,a3=b3=q^2=2,d=(a3-a1)2=1/2
2、N5-M5=b1+b2+b3+b4+b5-a1-a2-a3-a4-a5
因为a1=b1,a3=b3,a7=b5,代入上式，得
N5-M5=a1+b2+a3+b4+a7-a1-a2-a3-a4-a5=b2+b4+a5+2d-a2-a4-a5=b2+b4-a2-a4+2d
b2=√2,b4=2√2,2d=1,a2=3/2,a4=5/2,所以,N5-M5=√2+2√2+1-3/2-5/2=3√2-3>0,所以N5>M5
3、an=a1+(n-1)d=1+(n-1)*1/2=(n+1)*1/2,所以a(2n+1)=n+1
bn=q^(n-1),所以b(2n+1)=q^(2n+1-1)=q^(2n)=2^n
Cn=(n+1)*2^n设Cn前n项和为Sn，
Sn=2*2+3*2^2+4*2^3+-------------+(n+1)*2^n……①
nSn= 2*2^2+3*2^3+4*2^4+----------+n*2^n+(n+1)*2^(n+1)……②
①-②得，-Sn=4+2^2+2^3+2^4+--------+2^n-(n+1)*2^(n+1)
=2+2+2^2+2^3+2^4+----+2^n-(n+1)*2^(n+1)
=2+2*(1-2^n)/(1-2)-(n+1)*2^(n+1)
=2^(n+1)-(n+1)*2^(n+1)
-n*2^(n+1)
所以Sn=n*2^(n+1)