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Answer by Cesareo for solve recurrence relation $a_{n+2}=(a_{n+1}+1)/a_n$
Calling $a_1 = c_1$ and $a_2 = c_2$ we have$$\cases{a_1 = c_1\\a_2 = c_2\\a_3 = \frac{c_2+1}{c_1}\\a_4 = \frac{c_1+c_2+1}{c_1c_2}\\a_5 = \frac{c_1+1}{c_2}\\a_6 = c_1\\a_7 = c_2\\\vdots}$$
View ArticleAnswer by acreativename for solve recurrence relation $a_{n+2}=(a_{n+1}+1)/a_n$
Note $a_{1} = 2015, a_{2} = 2016, a_{3} = \frac{2017}{2015}, a_{4} = \frac{2}{2015},a_{5} = 1, a_{6} = 2015, a_{7} = 2016,...$Thus the sequence repeats modulo $5$. Hence $a_{2017} = a_{2} = 2016$.You...
View Articlesolve recurrence relation $a_{n+2}=(a_{n+1}+1)/a_n$
Let $a_1$ = 2015, $a_2$=2016. Find $a_{2017}$, given $a_{n+2}=(a_{n+1}+1)/a_n$I have found the 2 roots of the characteristic equation $x^2-x-1=0$ :$r_1 = ((1+\sqrt{5})/2)^n$ , $r_2 =...
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