Help please...
1) Use the method of mathematical induction to prove that is a multiple of 6 for n 1
2) For positive integers n and r, with r n, show that
where , do not use induction
This is non-sense. Having already said that 4^{k+1}+ 14= 6N adding the positive number 4^k+ 14 it won't still be "6N". In any case, your previous statement " 4^{k+1}+ 14= 6N" is not possible because you have already said that 4^k+ 14= 6N. The can't both be equal to the same number. You could say "4^{k+1}+ 14= 6M" for some "M", not necessarily "N". But that is what you want to prove- you cannot assert it yet.
Instead start with 4^{k+1}+ 14= 4(4^k)+ 14= 3(4^k)+ 4^k+ 14= 3(4^k)+ 6N.
Now use the fact that 4= 2^2 to write 4^k= (2^2)^k= 2^{2k} so that the above is 3(2)(2^{2k-1})+ 6N= 6(2^{2k-1}+ N)
The "tex" tags are not working so I took them out.
and for (2)
since ,
so