重酬!!!SUPER EASY MATHS (超趕ga)

1.
Assume n=k is true , i.e. 3^(4k+2) + 5^(2k+1) = 14N where N is some constant
When n=k+1
3^(4k+6) + 5^(2k+3)
=3^(4k+6) + 25( 5^(2k+1) )
=3^(4k+6) + 25 ( 14N - 3^(4k+2) )
= 14(25N)+ 3^(4k+2)( 81 - 25)
= 14[ 25N + 4x3^(4k+2) ]

2.
Assume n=k is true. i.e. 81x3^2k - 2^2k = 5N where N is some constant.
when n=k+1
81x3^(2k+2) - 2^(2k+2)
=81x3^(2k+2) - 4x2^2k
=81x3^(2k+2) - 4 ( 5N - 81x3^2k )
=81x3^2k( 9+1 ) - 20N
= 5 ( 162x3^2k - 4N )