M2 Mathematical Induction

Put n = 1,
=> 5(4*1-1)+1 = 16 ... It can be divided by 16
Put n = 2,
=> 25(4*2-1)+1 = 176 ... 176=11*16. It can be divided by 16
Assume that
P(n) = 5^n(4n-1)+1 can be divided by 16

Put n = n+1
=> 5^(n+1)[4(n+1)-1]+1
= 5^(n+1)(4n+3)+1
= 5*5^n(4n+3)+1
= 5*5^n(4n-1+4)+1+5-5
= 5*[5^n(4n-1)+1]+1+20-5
= 5*P(n)+16

Therefore, P(n+1) can also be devided by 16.
It is proved

= 5*5^n(4n-1+4)+1+5-5
= 5*[5^n(4n-1)+1]+1+20-5
?? wrong?? if right, pls. could show more steps?

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