I28-47

Question

The infinite sequence of nonzero integers  a_1 ,  a_2 ,  a_3 , ... is such that  a_n = (a_{n-1})(a_{n-2})^3  for  n > 2 , and  a_2 = 2 .
 
Select for  a_3  and  a_5  the values of  a_3  and  a_5 , respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.

Bunuel · Answer

Official Solution:

Given that  a_n = (a_{n-1})(a_{n-2})^3  and  a_2 = 2 , we have:
   
 a_{3} = (a_{2})(a_{1})^3 = 2(a_{1})^3 
 
 a_{4} = (a_{3})(a_{2})^3 = 8(a_{3}) 
 
 a_{5} = (a_{4})(a_{3})^3 
   
Substituting  a_{4} = 8(a_{3})  into  a_{5} = (a_{4})(a_{3})^3 , we get:
   
 a_{5} = 8(a_{3})(a_{3})^3 = 8(a_{3})^4 
   
So, we need such a pair of  a_3  and  a_5  that  a_{5} = 8(a_{3})^4 . Checking the options, only  a_3 = -2  and  a_5 = 128  work.

Correct answer:

a_3   "-2"

a_5   "128"

Karym · Answer

I like the solution - it’s helpful.

Helpmecat · Answer

I did not quite understand the solution. how do you get a3

Bunuel · Answer

a_3  is not determined directly, because  a_1  is unknown.

Using the recurrence:

a_3 = (a_2)(a_1)^3

a_3 = 2(a_1)^3

So  a_3  depends on the value of  a_1 . That is why we do not find  a_3  first by itself.

Instead, we express  a_5  in terms of  a_3 :

a_4 = (a_3)(a_2)^3 = 8a_3

a_5 = (a_4)(a_3)^3 = 8a_3(a_3)^3 = 8(a_3)^4

Now check the choices. Only  a_3 = -2  gives:

a_5 = 8(-2)^4 = 8 * 16 = 128

So the jointly consistent selections are  a_3 = -2  and  a_5 = 128 .

I28-47

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My Rewards

\(a_3\)	\(a_5\)
		128
		16
		8
		4
		-2
		-8

\(a_3\)	\(a_5\)
		128
		16
		8
		4
		-2
		-8