Question

在数列{an}中,已知a1=1/4,(an+1)/an=1/4,bn+2=3log1/4an(n∈N

在数列{an}中,已知a1=1/4,(an+1)/an=1/4,bn+2=3log1/4an(n∈N*)
设数列{cn}满足cn=an＋bn,求{cn}的前n项和Sn

Answer 1

an=(1/4)*(1/4)^(n-1)=(1/4)^n

bn+2=3log1/4an=3n
bn=3n-2
Sn=(1/4)+(1/4)^2+...+(1/4)^n+(1+4+....+3n-2)
=1/4*(1-(1/4)^n)/(1-1/4)+(1+3n-2)n/2
=1/3*(1-1/4^n)+(3n-1)n/2

Answer 2

Please clarify
1.(an+1)/an
a<n+1>/an or ((an) + 1)/an
2. bn+2=3log1/4an
bn+2 = b<n+2> or (bn) + 2
3log1/4an = 3log<1/4>an ( base 1/4) or 3log((1/4)an )

Answer 3

如果a(n+1)/an=1/4;b(n+2）=3log(1/4) an;
那么：Scn=San+Sbn=San+3nlog（1/4)*Sa(n-2)

Sn=a1(1-q^n)/(1-q), 这里面，a1=1/4,q=1/4，代入计算即可。