已知数列{an}是等差数列,{bn}是等比数列,在数列{cn}中,对任意n属于正整数,都有cn=an-bn,且c1=0,

数列{an}是等差数列,an=a1+(n-1)×d
{bn}是等比数列,bn=b1×q^(n-1)
c1=a1-b1=0,a1=b1
c2=a2-b2= a1+d -a1×q= 1/6 ,d+a1×(1-q) = 1/6 ①
c3=a3-b3 = a1+2d-a1×q^2= 2/9 ②
c4=a4-b4=a1+3d-a1×q^3= 7/54 ③
②-①,可得d+a1×q×(1-q) =1/18 ④
③-②,可得d+a1×q^2×(1-q) = -11/54 ⑤
①-④,可得a1×(1-q) ^2= 1/9 ⑥
④-⑤,可得a1×q×(1-q) ^2 = 7/27 ⑦
⑦/⑥,可得q= 7/3,将q=7/3代入⑥中,
可得a1= 1/16 =b1,将a1= 1/16和q= 7/3代入①中,
可得d= 1/4
所以,an=1/16 + (n-1)/4 = (4n-3)/16
{an}的前n项和= [(a1+an)×n]/2 = (2n^2-n)/16
bn=1/16×(7/3)^(n-1)
{bn}的前n项和= b1×(1-q^n)/(1-q)= 1/16 × 1/(7/3-1) × [(7/3)^n-1]
=3×[(7/3)^n-1]/64
数列{cn}的前n项和Sn={an}的前n项和 - {bn}的前n项和
=(2n^2-n)/16 - 3×[(7/3)^n-1]/64
={(8n^2-4n) - 3×[(7/3)^n-1]}/ 64