Prove by induction that 2^(6n)+3^(2n-2) is divsible by 5. (AS Further pure)

For this question I would heavily emphasise layout as these questions are very strict. Base n=1 simply plug n=1 into the equation and you will find you end up with 65 which is divisble by 5 Assumption n=k plug n=k into the equation. From here there are multiple different methods but I like to pause here and manipulate with the equation when in the form K as its just a method I like. Since we are assuming that this equation is divisble by 5 we can assume its a mulitple of 5. So I would write:2^6k+3^2k-2=5M. Manipulate the indicies by seperating them to end up with 2^6k+1/9(3^2k )=5M then multiply through by 9 to end up with 9(2^6k)+(3^2k )=45M. I would then re arrange to end up with 3^2k = 45M- 9(2^6k ). Then this would be where I would move on. Inductive step n=k+1 I would plug n=k+1 and after manipulating indicies as I demonstrated above i'd end up with 64(2^6k ) + 3^2k . However we have an equation now in the form 3^2k . I would then substitute this equation into the above formula ending up with 64(2^6k ) + 45M- 9(2^6k ).Then after final manipulation you end up with 55(2^6k ) + 45M. I would then write it as a multiple of 5 to demonstrate the proof is true.5 [11(2^6k ) + 9M] Conclusion The conclusion is practically always the same regardless of if we are proving by induction for series divison or matrices. The general layout is: If the summation is true for n=k and n=k+1 and shown to be true for n=1, then through the process of mathematical induction, the summation is true for all positive integers.