Question 1163040
<pre>
Something is wrong with this problem, as I show below.
</pre>
If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1.
<pre>
{{{matrix(1,5,

a[k] - 4a[k-1] + 3a[k-2] = 0, 
","  ,  
k >= 0, 
then,
a[k] = (1 + p)/q 

)}}}

We substitute k=2

{{{matrix(1,5,

a[2] - 4a[2-1] + 3a[2-2] = 0, 
","  ,  
2 >= 0, 
so,
a[2] = (1 + p)/q 

)}}}

{{{a[2] - 4a[1] + 3a[0] = 0}}}

{{{2 - 4(1) + 3a[0] = 0}}}

{{{2 - 4 + 3a[0] = 0}}}

{{{matrix(1,5,

-2 + 3a[0] = 0, 
""  ,  
"", 
"",
""
)}}}

{{{matrix(1,5,

3a[0] = 2, 
""  ,  
"", 
so,
2 = (1 + p)/q 

)}}}

{{{matrix(1,5,

a[0] = 2/3, 
""  ,  
"", 
so,
2q = 1 + p 

)}}}

-----

Next we substitute k=3

{{{matrix(1,5,

a[3] - 4a[3-1] + 3a[3-2] = 0, 
","  ,  
2 >= 0, 
so,
a[3] = (1 + p)/q 

)}}}

{{{a[3] - 4a[2] + 3a[1] = 0}}}

{{{a[3] - 4(2) + 3(1) = 0}}}

{{{a[3] - 8 + 3 = 0}}}

{{{matrix(1,5,

a[3] -5 = 0, 
""  ,  
"", 
"",
""
)}}}

{{{matrix(1,5,

a[3] = 5, 
""  ,  
"", 
so,
5 = (1 + p)/q 

)}}}

{{{matrix(1,5,

a[3] = 5, 
""  ,  
"", 
so,
5q = 1 + p 

)}}}

So we have these two equations in p and q:

{{{system(2q=1+p,5q=1+p)}}}

This can't be because the solution to that system is p=-1, q=0,

But q is in the denominator of 

{{{a[k] = (1 + p)/q}}}

and so q cannot be 0.  So the problem is botched.

You can correct it in the space below if you like
and I'll get back to you by email. 

Edwin</pre>