mathematical Induction (MI) problem

Let S(k) be the statement 5^k(4k-1)+1 is divisible by 16
When k=1
5^1[4(1)-1]+1=16
Therefore S(1) is true.
Assume that S(n) is true
When k=n+1
5^k(4k-1)+1
=5^(n+1) [4(n+1)-1]
=5(5^n)[4(n+1)]-5(5^n)+1
=20(5^n)(n+1)-5(5^n)+1
=[20(n+1)-5]5^n+1
=5^(n+1)[4(n+1)-1]+1
=5^k(4k-1)+1
Since S(n) is true,S(n+1) is true.
Therefore by the priciple of mathematical induction, S(k) is true for all natural number k.

when k = 1,

(5^k)(4k-1)+1
= (5)(3) + 1
= 16

therefore the proposition is true when k = 1

Assume the propsition is true for k = m

(5^m)(4m-1)+1 = 16b, for some positive interger b

When k = m+1

[5^(m+1)][4(m+1)-1]+1
= (5)(5^m)[4m+4-1]+1
= (5)(5^m)(4m-1) + (4)(5)(5^m) + 1
= (5)(5^m)(4m-1) + (4)(5)(5^m) + 1
= (5)(5^m)(4m-1) + 5 + (4)(5)(5^m) - 4
= (5)(16b) + (4)(5)(5^m) - 4
= 16 (5b + {[5^(m+1)] -1}/4)
= (16)(5){b+[(5^m)-1]/4}

As for all positive interger m, [(5^m)-1] is divisible by 4 (you may prove it by M.I.), {b+[(5^m)-1]/4} is an integer

Therefore (16)(5){b+[(5^m)-1]/4} is an integer and is divisible by 16.

Thus the proposition is true for k=m+1

By M.I., the proposition is true for all positive integer k