The discussion focuses on solving the recurrence relation defined by the equation $(a_{n+1}+3)(2-a_n)=6$ with the initial condition $a_1=1$. By transforming the sequence into $x_n = \frac{1}{a_n}$, the relation simplifies to $x_{n+1} = \frac{1}{3}(2x_n - 1)$, yielding the closed form $x_n = 2\left(\frac{2}{3}\right)^{n-1} - 1$. The sum $\sum_{n=1}^{100} x_n$ is calculated as $6\left(1 - \left(\frac{2}{3}\right)^{100}\right) - 100$, resulting in an approximate value of $-94$, accurate to 17 decimal places.
PREREQUISITESMathematicians, students studying discrete mathematics, and anyone interested in solving recurrence relations and series summation problems.
[sp]If $(a_{n+1}+3)(2-a_n)=6$ then $2a_{n+1} - 3a_n - a_{n+1}a_n = 0$ and so $$\frac2{a_n} - \frac3{a_{n+1}} = 1.$$Albert said:given :$(a_{n+1}+3)(2-a_n)=6$
$(a_n\neq 0, a_1=1)$
please find :$\sum_{n=1}^{100} \dfrac {1}{a_n}$
perfect!Opalg said:[sp]If $(a_{n+1}+3)(2-a_n)=6$ then $2a_{n+1} - 3a_n - a_{n+1}a_n = 0$ and so $$\frac2{a_n} - \frac3{a_{n+1}} = 1.$$Let $x_n = \dfrac1{a_n}$. Then $x_{n+1} = \frac13(2x_n - 1)$, with $x_1 = 1$. That is a linear recurrence relation, with solution $x_n = 2\bigl(\frac23\bigr)^{\!n-1} - 1$. Therefore $$\sum_{n=1}^{100}x_n = \sum_{n=1}^{100} \Bigl( 2\bigl(\tfrac23\bigr)^{\!n-1} - 1 \Bigr) = \frac{2\Bigl( 1 - \bigl(\frac23\bigr)^{\!100} \Bigr)}{1-\frac23} - 100 = 6\Bigl(1 - \bigl(\tfrac23\bigr)^{\!100} \Bigr) - 100.$$The approximate solution $$\sum_{n=1}^{100}x_n \approx -94$$ is correct to about 17 decimal places.[/sp]