Just out of interest, why did you post this in the Geometry forum?
Anyway, can you at least prove the base case?
Use mathematical induction to prove that
4 x 6^(2n) + 3 x 2^(3n) is divisible by 7 for n=1,2,3...
I roughly know the method but I can't seem to get there with this. Please could someone take the time to post a detailed answer? Thank you
I wasn't sure which forum it should be in...where would you suggest I post it?
And yes, I have proved the base case, and I know I have to assume that p(k) is true and use this to prove p(k+1) but I am unsure of how to proceed from there on.
I have p(k) is 4 x 6^(2k) + 3 x 2^(3k) = 7m where m is an integer.
And I have to use that to show that p(k+1) is divisible by 7.
So I have 4 x 6^2(k+1) + 3 x 2^3(k+1)
So far I have written it as 4 x 6^(2k+2) + 3 x 2^(3k+3)
And from there I have split the powers to get
4 x 6 x 6^(2k+1) +3 x 4 x 2^(3k+1) ... this is the line that I am unsure of and I don't know if it's right or not. If it is, I may need some guidance as to where to go from here.
We are trying to reduce the expression for the case of k + 1 to either the case for k or a combination of the cases for 1 and k.
I have found that it helps sometimes in finding a proof to actually work out the cases for n = 1, n = 2, n = 3. You have to do that for n = 1 to do a proof by induction anyway.
$4 * 6^{(2 * 1)} + 3 * 2^{(3 * 1)} = 4 * 36 + 3 * 8 = 144 + 24 = 168 = 7 * 24.$
$4 * 6^{(2 * 2)} + 3 * 2^{(3 * 2)} = 4 * 6^4 + 3 * 2^6 = 4 * 1296 + 3 * 64 = 5284 + 192 = 5376 = 7 * 768.$
$4 * 6^{(2 * 3)} + 3 * 2^{(3 * 3)} = 4 * 6^6 + 3 * 2^9 = 4 * 46656 + 3 * 512 = 186624 + 1536 = 188160 = 7 * 26680.$
Now that is initially disappointing because I at least do not see an obvious relationship between the various multiples of 7. But I now have some confidence that what I am trying to prove may actually be true.
Now I play around a bit 5376 - 168 = 5208 = 7 * 744. And 188160 - 5376 = 182784 = 7 * 26112. Well of course, the difference between two multiples of 7 is itself a multiple of 7. Maybe that will give me a proof.
$k \in \mathbb Z\ and\ k > 0\ and\ 4 * 6^{2k} + 3 * 2^{3k} = 7m_k,\ where\ m_k \in \mathbb Z\ and\ m_k > 0.$
$4 * 6^{2(k + 1)} + 3 * 2^{3(k + 1)} = 4 * 6^{2k} * 6^2 + 3 * 2^{3k} * 2^3 = 36(4 * 6^{2k}) + 8(3 * 2^{3k}) .$
$So\ 4 * 6^{2(k + 1)} + 3 * 2^{3(k + 1)} - (4 * 6^{2k} + 3 * 2^{3k}) =$
$36(4 * 6^{2k}) + 8(3 * 2^{3k}) - (4 * 6^{2k} + 3 * 2^{3k}) =$
$35(4 * 6^{2k}) + 7(3 * 2^{3k}) = 7(20 * 6^{2k} + 3 * 2^{3k}).$
$But\ obviously\ 20 * 6^{2k} + 3 * 2^{3k}\ is\ an\ integer > 0\ because\ k\ is\ a\ positive\ integer.$
$So\ \exists\ j_{k + 1} \in \mathbb Z\ such\ that\ j_{k + 1} = 20 * 6^{2k} + 3 * 2^{3k} > 0.$
$\therefore 4 * 6^{2(k + 1)} + 3 * 2^{3(k + 1)} - (4 * 6^{2k} + 3 * 2^{3k}) = 7j_{k + 1} \implies$
$4 * 6^{2(k + 1)} + 3 * 2^{3(k + 1)} = 7j_{k + 1} + 4 * 6^{2k} + 3 * 2^{3k} = 7j_{k + 1} + 7m_k = 7(j_{k + 1} + m_k).$
There is no guaranteed route to a proof. You just have to try various things.