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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?

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caylus
Hello,

[tex]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* \frac{r^3-1}{r-1} \\ ...\\ u_{n}=r*u_{n-1}+s\\ \ =(r^{n-1}*u_{1}+s* \frac{r^{n-1}-1}{r-1})*r+s \\ \ =r^{n}*u_{1}+s* \frac{r^n-r+r-1}{r-1} \\ \boxed{u_{n}=r^{n}*u_{1}+s* \dfrac{r^n-1}{r-1}} \\ \\ \ u_{56}=2^{56}*28-15* \frac{2^{56}-1}{2-1} =936748722493063183 \\ [/tex]

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