Sum of Sequence $(a_{n+1}+3)(2-a_n)=6$ to 100

[Albert1]

given :$(a_{n+1}+3)(2-a_n)=6$
$(a_n\neq 0, a_1=1)$
please find :$\sum_{n=1}^{100} \dfrac {1}{a_n}$

 

[Opalg]

given :$(a_{n+1}+3)(2-a_n)=6$
$(a_n\neq 0, a_1=1)$
please find :$\sum_{n=1}^{100} \dfrac {1}{a_n}$

[sp]If $(a_{n+1}+3)(2-a_n)=6$ then $2a_{n+1} - 3a_n - a_{n+1}a_n = 0$ and so \(\displaystyle \frac2{a_n} - \frac3{a_{n+1}} = 1.\)

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 \(\displaystyle \sum_{n=1}^{100}x_n \approx -94\) is correct to about 17 decimal places.[/sp]

 

[Albert1]

[sp]If $(a_{n+1}+3)(2-a_n)=6$ then $2a_{n+1} - 3a_n - a_{n+1}a_n = 0$ and so \(\displaystyle \frac2{a_n} - \frac3{a_{n+1}} = 1.\)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 \(\displaystyle \sum_{n=1}^{100}x_n \approx -94\) is correct to about 17 decimal places.[/sp]

perfect!