Question

How would someone write a shifted geometric sequence in the explicit formula. I believe it requires modification of f(x) = ln(x) however, I haven't been able to figure it out. For example U(n) = U(n-1)(r)-15, let us say U(1) = 28 and r = 2, how would I figure out the 56th term without a graphing calculator using the explicit formula?

Answer 1

Hello,


`r=2\\ s=-15\\ u_(1)=28\\ u_(n)=r*u_(n-1) +s\\ u_(2)=r*u_(1)+s\\ u_(3)=r*u_(2)+s=r^2*u_(1)+r*s+s\\ u_(4)=r*u_(3)+s\\ =r^3*u_(1)+r^(2)*s+r*s+s\\ =r^3*u_(1)+s(r^2+r+1)=r^3*u_(1)+s* (r^3-1)/(r-1) \\ ...\\ u_(n)=r*u_(n-1)+s\\ \ =(r^(n-1)*u_(1)+s* (r^(n-1)-1)/(r-1))*r+s \\ \ =r^(n)*u_(1)+s* (r^n-r+r-1)/(r-1) \\ \boxed{u_(n)=r^(n)*u_(1)+s* (r^n-1)/(r-1)} \\ \\ \ u_(56)=2^(56)*28-15* (2^(56)-1)/(2-1) =936748722493063183 \\`