Question

数列an的通项公式an=6n-5（n为奇数）,an=2的n次方（n为偶数）,求数列an的前n项的和Sn

Answer 1

1.

当n为偶数时,n=2k

a(2k-1)=6(2k-1)-5)=12k-11

sk=12k(k+1)/2-11k=6k^2-5k

a(2k)=2^(2k)=4^k

tk=4(4^k-1)/3

=(1/3)4^(k+1)-4/3

S(2k)=sk+tk=6k^2-5k+4(4^k-1)/3

=6k^2-5k+(1/3)4^(k+1)-4/3

=(1/3)4^(k+1)+6k^2-5k-4/3

Sn=(1/3)4^(n/2+1)+6(n/2)^2-5n/2-4/3

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

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

2.

当n为奇数时,n=2k-1

a(2k-1)=6(2k-1)-5=12k-11

sk=12k(k+1)/2-11k

=6k^2-5k

a(2k-2)=2^(2k-2)=4^(k-1)

tk=4[4^(k-1)-1]/3

=(1/3)4^k-4/3

S(2k-1)=sk+tk

=6k^2-5k+(1/3)4^k-4/3

=(1/3)4^k+6k^2-5k-4/3

Sn=(1/3)4^[(n+1)/2]+6[(n+1)/2]^2-5(n+1)/2-4/3

=(1/3)2^(n+1)+(3/2)(n+1)^2-(5/2)n-23/6

综上所述

当n为偶数时,Sn=(1/3)2^(n+2)+(3/2)n^2-(5/2)n-4/3

当n为奇数时,Sn=(1/3)2^(n+1)+(3/2)(n+1)^2-(5/2)n-23/6