Question

bn=2n-1,an=(1/2)^(n-1)若cn=bn/an求cn的前n项和tn

Answer 1

cn=bn/an=(2n-1)/(1/2)^(n-1)=(2n-1)(2)^(n-1)

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

2Tn=1*2^1+3*2^2+5*2^3+……+(2n-1)*2^(n)

-Tn=Tn-2Tn=1*2^0+2*2^1+2*2^2+……+2*2^(n-1)-(2n-1)*2^(n)

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

=1+2(2^n-2)-(2n-1)*2^(n)

=2*2^n-(2n-1)*2^(n)-3

Tn=2^(n)×(2n-3)+3

知识点：类似等差乘以等比的数列求前N项和,只需乘以等比数列的公比,再错位相减,则得到的新的等比数列和某几项特殊项.