已知定义在R上的函数f(x)是奇函数且满足f(3/2-x)=f(x),f(-2)=-3,数列{an}满足a1=-1,且Sn=2an+n,则f(a5)+f(a6)=?
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已知定义在R上的函数f(x)是奇函数且满足f(3/2-x)=f(x),f(-2)=-3,数列{an}满足a1=-1,且Sn=2an+n,则f(a5)+f(a6)=?
已知定义在R上的函数f(x)是奇函数且满足f(3/2-x)=f(x),f(-2)=-3,数列{an}满足a1=-1,且Sn=2an+n,则f(a5)+f(a6)=?
已知定义在R上的函数f(x)是奇函数且满足f(3/2-x)=f(x),f(-2)=-3,数列{an}满足a1=-1,且Sn=2an+n,则f(a5)+f(a6)=?
an + S(n-1) = Sn = 2an+n
an = S(n-1) - n
a1 = -1,Sn = -1
a2 = S1 - 2 = -3,S2 = - 4
a3 = S2 - 3 = -7,S3 = -11
a4 = S3 - 4 = -15,S4 = -26
a5 = S4 - 5 = -31,S5 = -57
a6 = S5 - 6 = -63
f(3/2-x) = f(x)
f(x) = f(3/2-x) = -f(x - 3/2) = - f(3-x) = f(x -3)
所以同期为3
f(a5) = f(-31) = f(2) = -f(2) = 3;
f(a6) = f(-63) = f(0) = 0
f(a5)+fa6) =3
通过f(3/2-x)=f(x)，f(-2)=-3求出f(x)的表达式，在通过a1=-1，且Sn=2an+n，求出a5，a6多少代入f(a5)+f(a6)=？就可以知道答案
s（n-1）=2a（n-1）+(n-1)
sn-s(n-1)=2(an-a(n-1))+1=an,则an=2a(n-1)-1
利用迭代法an=2^(n-1)*a1-(1+2+...+2^(n-2))=1-2^n
a5=-31,a6=-63
f(3/2-x)=f(x)=-f(-x).则f(3/2+x)=-f(x)=-f(3+x)
周期为3且f(0)=0,f(-2)=-3
f(a5)=f(-31)=f(-2)=-3
f(a60=f(-63)=f(0)=0
结果为-3
WQEEDEDED