AS further

[bright hinge]

Can someone help me on question 3 please ?

[mental viper]

first bit is the part where for some value of k, its divisible by 5 (so we know this is a variable divisible by 5)
second part is where the subbing part of k+1 happens
in the final part you can factor out the 2^6 from the original , getting a remainder of 55*3^2k-2 (a multiple of 5)
since both parts are divisible by 5, the k + 1 substitution is also divisible by 5

i find a good rule of thumb is to try to factor out the original k substitution, so that you often get a remainder which should also follow the requirement of the question (like divisible by 5 here)

[mossy laurel]

I love this technique

Whatever you have like in this case it’s raised to 2k+6 and the original is ^2k so remove 2^6 from it and do the same to the other

If it was

2^k and 3^k+1

And after you do f(k+1) it’s 2^k+1 and 3^k+2 so you’ll remove 2^1 and 3^1

[bright hinge]

Yeah thank you I’m still kind of on the fence but I’ll get there thank you very much !

[bright hinge]

I watxhed a few videos I understand it now