Given that $\mathrm{f}(r)=\frac{r+1}{{{3}^{r-1}}}$ where $r\ge 0$, show that \(\mathrm

Question

Given that $\mathrm{f}(r)=\frac{r+1}{{{3}^{r-1}}}$ where $r\ge 0$, show that $\mathrm{f}(r)-\mathrm{f}(r+1)=\frac{2r+1}{{{3}^{r}}}$[1] Use the above result to find $\frac{1}{3}+\frac{3}{{{3}^{2}}}+\frac{5}{{{3}^{3}}}+...+\frac{2n-1}{{{3}^{n}}}$, giving your answer in the form $A-\frac{Bn+C}{{{3}^{n}}}$, where $A,B$ and $C$ are constants to be determined. Hence find $\sum\limits_{r=n+1}^{\infty }{\frac{2r-1}{{{3}^{r}}}}$ in terms of $n$, justifying your answer.[9]

Accepted Answer

$\dfrac{n+1}{3^n}$

2020 J2 CT NYJC P1 Q7

2020 J2 CT NYJC P1 Q7

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