Question

已知数列{an}满足a1=0,a2=2,且对任意m'n属于N*,都有a(2m-1)+a(2n+1)=2a(m+n-1)+2(m-n)^2

设cn=(a(n+1)-an)q^(n-1),求数列{cn}的前n项和Sn

Answer 1

a(2n-1)+a(2n+1)=2a(n+n-1)+0=2a(2n-1),
a(2n+1)=a(2n-1)=...=a(1)=0.
a(2n-1)=0.
a[2(n+1)-1] + a(2n+1) = 2a(n+1 + n-1) + 2(n+1-n)^2,
a(2n+1) + a(2n+1) = 0 = 2a(2n) + 2,
a(2n)=-1, 与a(2)=2矛盾哈.
题目有问题吧...

追问

题目就是这样的！！！我也不会做

追答

若不顾问题的话,解法如下:
m=n+1,
a[2(n+1)-1] + a(2n+1)=2a(n+1+n-1)+2(n+1-n)^2,
2a(2n+1)=2a(2n) + 2,
a(2n+1)=a(2n)+1,
c(2n)=[a(2n+1)-a(2n)]q^(2n-1) = q^(2n-1),

m=n,
a(2n-1)+a(2n+1)=2a(n+n-1)+0=2a(2n-1),
a(2n+1)=a(2n-1)
a(2n-1)=a(2n+1)=a(2n)+1,
c(2n-1)=[a(2n)-a(2n-1)]q^(2n-2)=-q^(2n-2).
c(2n-1)+c(2n)=q^(2n-1)-q^(2n-2)=(q-1)q^(2n-2)=(q-1)(q^2)^(n-1)
若q^2=1,
s(2n)=[c(1)+c(2)]+...+[c(2n-1)+c(2n)]=[q-1] + [q-1]q^2 + ... + (q-1)(q^2)^(n-1)
=(q-1)n
s(2n-1)=s(2n)-c(2n)=(q-1)n-q^(2n-1).
若q^2不等于1,
s(2n)=[c(1)+c(2)]+...+[c(2n-1)+c(2n)]=[q-1] + [q-1]q^2 + ... + (q-1)(q^2)^(n-1)
=(q-1)[(q^2)^n-1]/(q^2-1)=[q^(2n)-1]/(q+1)
s(2n-1)=s(2n)-c(2n)=[q^(2n)-1]/(q+1)-q^(2n-1).