Mathematical induction~20marks

Prove that 3^(2n) - 1 is divisible by 8 is true

Solution:

when n=1 , 3^(2)-1=8 , so it is true for n=1

Assume it is true for n=k , where k is a nature number ,

i.e. 3^(2k)-1=8M, where M is a positive integer.

Consider n =k+1 ,

3^[2(k+1)]-1=3^(2k) 3^(2)-1

=9(8M+1)-1 ( By assumption)

=8(9M+1) it is true for n=k+1

By mathematical induction , it is true for all positive integer n.

Prove that (10^n) + 3(4^n) – 4 is divisible by 9 is true

Solution :

When n=1 , 10^1+3(4^1)-4=10+12-4=18=2x9

so it is ture for n=1

Assume it is true for k , where k is a positive integer ,

i.e. 10^k+3(4^k)-4=9N , where N is a positive integer ,

Consider n=k+1 ,

10^(k+1)+3(4^(k+1))-4

=10(9N+4-3(4^k)-10^k)+3(4^(k+1))-4

=9(10N)+(40-4)-3(4^k)(4-10)

=9(10N+6-2(4^k)) it is true for n=k+1

By Mathematic induction , it is true for all +ve integer n.

Prove that (5^n) (4n – 1) + 1 is divisible by 16

When n=1 , 5^1 (4-1)+1=16

it is true for n=1 ,

Assume it is true for n=k , where k is a +ve integer ,

i.e. 5^k(4k-1)+1=16Q ,

Consider n=k+1 ,

5^(k+1)(4k+3)+1

=5x5^(k) (4k-1) +4x5^(k+1)+1

=5(16Q-1)+4x5^(k+1)+1

=16x5!-5+1+4x5^(k+1)

=16x5+4(5^(k+1)-1)

P.s. Prove 5^(k+1)-1 = 4M by your self ,

=16x5+16C ,

so it is true for n=K+1

By Mathematic induction , it is true for all +ve integer n.