Return to Answer

Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $a\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $a\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.

There are, of course, other answers, of which I will show later.

This is another possible solution:

If this is of the form $a_n = b_n, a_n ∈ A. b_n ∈ B$): $A = \{2,3,4,5,6,9\}$ and $B = \{x_n ∈ A : x_nx_{n-1}\}$. Since $a_n = b_n$, we must find $a_6$, which is $b_6$. $b_6 = a_6a_5 = 9(6) = 54$.

I thus also conclude $9 = 54$.

(As a side note, using the "equality" sign to represent some function sort of rubs me the wrong way, as it has a long-standing representation. I do like how here (as opposed to how I originally saw it), the ‘→’ is used instead. Kudos!)

See here for a discussion, from which I drew my former answer (from myself). The former solution was just another way of tackling the common answer, also from myself.

Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $n\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $n\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.

There are, of course, other answers, of which I will show later.

This is another possible solution:

If this is of the form $a_n = b_n, a_n ∈ A. b_n ∈ B$): $A = \{2,3,4,5,6,9\}$ and $B = \{x_n ∈ A : x_nx_{n-1}\}$. Since $a_n = b_n$, we must find $a_6$, which is $b_6$. $b_6 = a_6a_5 = 9(6) = 54$.

I thus also conclude $9 = 54$.

(As a side note, using the "equality" sign to represent some function sort of rubs me the wrong way, as it has a long-standing representation. I do like how here (as opposed to how I originally saw it), the ‘→’ is used instead. Kudos!)

See here for a discussion, from which I drew my former answer (from myself). The former solution was just another way of tackling the common answer, also from myself.

Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $a\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $a\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.

There are, of course, other answers, of which I will show later.

Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $a\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $a\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.

There are, of course, other answers, of which I will show later.

This is another possible solution:

If this is of the form $a_n = b_n, a_n ∈ A. b_n ∈ B$): $A = \{2,3,4,5,6,9\}$ and $B = \{x_n ∈ A : x_nx_{n-1}\}$. Since $a_n = b_n$, we must find $a_6$, which is $b_6$. $b_6 = a_6a_5 = 9(6) = 54$.

I thus also conclude $9 = 54$.

(As a side note, using the "equality" sign to represent some function sort of rubs me the wrong way, as it has a long-standing representation. I do like how here (as opposed to how I originally saw it), the ‘→’ is used instead. Kudos!)

See here for a discussion, from which I drew my former answer (from myself). The former solution was just another way of tackling the common answer, also from myself.

Obviously, this follows the pattern $b_n=2n +b_{n-1}$. $a\in A,b\in B;A=\{2,3,4,5,6,9\}.$ So $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$

There are, of course, other answers, of which I will show later.

Obviously, this follows the pattern $b_n=2n +b_{n-1}$ for $a\in\{3,4,5,6\}$, with $b_2=6$. Assuming the pattern continues for $a\in\{7,8,9\}$, we get $b_9=2(9)+b_8=18+2(8)+b_7=34+2(7)+b_6=48+42=90$.

There are, of course, other answers, of which I will show later.