Question

f(n+1) = -2f(n) + 8 and f(1)= 1 for n>1

Answer 1

So whats the question bro?

Answer 2

observe that $$f(n+1)=-2f(n)+8\\f(n+2)=-2f(n+1)+8=-2(-2f(n)+8)+8=(-2)^2f(n)+8(1+(-2))\\f(n+k)=(-2)^kf(n)+8(1+(-2)+\dots+(-2)^{k-1})\\f(n+k)=(-2)^kf(n)+8\cdot\frac{1-(-2)^k}{1-(-2)}=(-2)^kf(n)+\frac83(1-(-2)^k)$$ergo \(f(n)=(-2)^{n-1}f(1)+\dfrac83(1-(-2)^{n-1})=(-2)^{n-1}+\frac83(1-(-2)^{n-1})\) so our final solution is $$f(n)=-\frac53(-2)^{n-1}+\frac83$$