Ummm... what is this?

Tia2007 wrote: Ummm... what is this? Im don't understand ur question

JAAAHS12 wrote: The infinity sequence $r_{1},r_{2},...$ of rational numbers it satisfies that: $\prod_{i=1}^ {k}r_{i}=\sum_{i=1}^{k} r_{i}$. For all natural k. Show that $\frac{1}{r_{n}}-\frac{3}{4}$ is a square of rationale number for all natural $n\geq3$ I suppose that sequence is considered as always nonzero (else $\frac 1{r_n}$ is meaningless). Let $S_n=\prod_1^nr_i=\sum_1^nr_i$ We get $S_nr_{n+1}=S_n+r_{n+1}$ and so $r_{n+1}=\frac{S_n}{S_n-1}$ And so $S_{n+1}=\frac{S_n^2}{S_n-1}$ So $\frac 1{r_{n+2}}=1-\frac 1{S_{n+1}}=1-\frac 1{S_n}+\frac 1{S_n^2}$ And so $\frac 1{r_{n+2}}-\frac 34=\frac 14-\frac 1{S_n}+\frac 1{S_n^2}=\left(\frac 1{S_n}-\frac 12\right)^2$ Q.E.D.

pco wrote: JAAAHS12 wrote: The infinity sequence $r_{1},r_{2},...$ of rational numbers it satisfies that: $\prod_{i=1}^ {k}r_{i}=\sum_{i=1}^{k} r_{i}$. For all natural k. Show that $\frac{1}{r_{n}}-\frac{3}{4}$ is a square of rationale number for all natural $n\geq3$ I suppose that sequence is considered as always nonzero (else $\frac 1{r_n}$ is meaningless). Let $S_n=\prod_1^nr_i=\sum_1^nr_i$ We get $S_nr_{n+1}=S_n+r_{n+1}$ and so $r_{n+1}=\frac{S_n}{S_n-1}$ And so $S_{n+1}=\frac{S_n^2}{S_n-1}$ So $\frac 1{r_{n+2}}=1-\frac 1{S_{n+1}}=1-\frac 1{S_n}+\frac 1{S_n^2}$ And so $\frac 1{r_{n+2}}-\frac 34=\frac 14-\frac 1{S_n}+\frac 1{S_n^2}=\left(\frac 1{S_n}-\frac 12\right)^2$ Q.E.D. U can share ur solution for?: https://artofproblemsolving.com/community/u446058h1737288p11282974

Oh sorry I was actually replying to another topic XD