Question

数列{an}中，a1=-27，an+1+an=3n-54，求数列{an}的通项公式 急需~

Answer 1

a(n+1)=-an+3n-54
a(n+1)+x(n+1)+y=-an+3n-54+x(n+1)+y
a(n+1)+x(n+1)+y=-[an-(3+x)n+54-x-y]
令x=-(3+x)
y=54-x-y
x=-3/2,y=111/4
a(n+1)-3/2(n+1)+111/4=-(an-3/2*n+111/4)
[a(n+1)-3/2(n+1)+111/4]/(an-3/2*n+111/4)=-1
所以an-3/2*n+111/4是等比数列，q=-1
所以an-3/2*n+111/4=(a1-3/2*1+111/4)*(-1)^(n-1)=-3/4*(-1)^(n-1)
an=-3/4*(-1)^(n-1)+3/2*n-111/4

Answer 2

利用恒等式sin²x+cos²x=1
cos²B+cos²C=sin²A(cos²θ+sin²θ)
(1-sin²B)+(1-sin²C)=sin²A
所以sin²A+sin²B+sin²C=2