已知定义在R上的函数f(x)是奇函数且满足f(3
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}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+nan = S(n-1) - na1 = -1, Sn = -1a2 = S1 - 2 = -3, S2 = - 4a3 = S2 - 3 = -7, S3 = -11a4 = S3 - 4 = -15, S4 = -26a5 = S4 - 5 = -31, S5 = -57a6 = S5 - 6 = -63f(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3f(a5) = f(-31) = f(2) = -f(2) = 3;f(a6) = f(-63) = f(0) = 0f(a5)+fa6) =3】】【【f(x) = f(3/2-x) = -f(x - 3/2) = - f(3-x) = f(x -3)所以同期为3f(a5) = f(-31) = f(2) = -f(2) = 3;f(a6) = f(-63) = f(0) = 0f(a5)+fa6) =3】】这里不明白。
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