1.

Using mathematical induction prove that 2^3n -1 is divisible by 7​

Answer»

To prove :

2^(3n) - 1 is divisible by 7

Proof :

Let P(n) : 2^(3n) - 1 is divisible by 7 be a statement .

If n = 1 , then

2^(3n) - 1 = 2^(3•1) - 1

= 2³ - 1

= 8 - 1

= 7 (which is divisible by 7)

Clearly ,

For n = 1 , 2^(3n) - 1 is divisible by 7 .

Thus ,

P(1) is true .

Now ,

Let us ASSUME that P(k) , k € N is true .

ie .

→ 2^(3k) - 1 is divisible by 7

→ 2^(3k) - 1 = 7m , m € N -----(1)

Now ,

If n = k + 1 , then

→ 2^[3(k+1)] - 1

= 2^(3k + 3) - 1

= 2^3k • 2³ - 1

= (7m + 1)•2³ - 1 [ using EQ-(1) ]

= (7m + 1)•8 - 1

= 7•8m + 8 - 1

= 7•8m + 7

= 7•(8m + 1)

= 7P where p = (8m + 1) € N (divisible by 7)

Clearly ,

For n = k + 1 , 2^(3n) - 1 is divisible by 7 .

Thus ,

P(k + 1) is true .

Since ,

P(k + 1) is true whenever P(k) is true , thus

P(n) is true for all n € N .

Hence ,

2^3n - 1 is divisible by 7 .

Hence proved .


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