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)=?](/uploads/image/z/8821133-53-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%E4%B8%94%E6%BB%A1%E8%B6%B3f%283%2F2-x%29%3Df%28x%29%2Cf%28-2%29%3D-3%2C%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D-1%2C%E4%B8%94Sn%3D2an%2Bn%2C%E5%88%99f%28a5%29%2Bf%28a6%29%3D%3F)
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