Question

问一道高二数学题（与数列有关）已知等差数列{an}的通项公式为an=4n-2,等比数列{bn}的通项公式为bn=2/4^

Answer 1

cn=an/bn

=(4n-2)/[2/4^(n-1)]

=(2n-1)4^(n-1)

cn=(2n-1)4^(n-1)

Pn=(2n-1)4^(n-1)+(2n-3)4^(n-2)+(2n-5)4^(n-3)+……+5*4^2+3*4^1+4^0

4Pn=(2n-1)4^n+(2n-3)4^(n-1)+(2n-5)4^(n-2)+……+5*4^3+3*4^2+4^1

两式相减：

-3Pn=-(2n-1)4^n+2*4^(n-1)+2*4^(n-2)+……+2*4^3+2*4^2+2*4^1+1

=-(2n-1)4^n+2[4^(n-1)+4^(n-2)+……+4^3+4^2+4^1]+1

=-(2n-1)4^n+2*4[4^(n-1)-1]/(4-1)+1

=-(2n-1)4^n+(8/3)*4^(n-1)-5/3

=-(2n-1)4^n+(2/3)*4^n-5/3

=-(2n-5/3)4^n-5/3

Pn=(1/9)(6n-5)4^n+5/9