M.I.題目唔識做

Let P(n) be the statement (6n - 1) x 7n + 1 is divisible by 36

When n = 1, (6n - 1) x 7n + 1 = 36 which is divisible by 36.

So P(1) is true.

Assume P(k) is true where k is a positive integer, i.e.

(6k - 1) x 7k + 1 = 36N where N is a natural no.

When n = k + 1, the expression becomes:

(6k + 5) x 7k+1 + 1

= 7 x (6k + 5) x 7k + 1

= 7 x [(6k - 1) + 6] x 7k + 1

= 7 x (6k - 1) x 7k + 42 x 7k + 1

= 7 x (6k - 1) x 7k + 7 + 42 x 7k - 6

= 7 x [(6k - 1) x 7k + 1] + 6 x (7k - 1)

= 7 x 36N + 6 x (7 - 1) x (7k + 7k-1 + 7k-2 + ... + 1)

= 7 x 36N + 36 x (7k + 7k-1 + 7k-2 + ... + 1)

So the expression is also divisible by 36 since 7k + 7k-1 + 7k-2 + ... + 1 is a positive integer when k is a positive integer.

Thus P(k + 1) is also true and by the principle of MI, P(n) is true for all positive integers n.

2010-10-23 13:58:26 補充：
Note: All "7k" refer to "7 to the powers of k", so do "7n" and "7k-1" and "7k-2", etc.

Let P(n) be (6n - 1) x 7n + 1 is divisible by 36 for all positive integers n

When n = 1, (6 - 1) x 7 + 1 = 36 which is divisible by 36.
.'.P(1) is true.

Assume P(k) is true where k is a positive integer,
i.e.
(6k - 1) x 7k + 1 = 36N where N is a integer

When n = k + 1,
(6k +6 -1) x 7k+1 + 1
= 7 x [(6k - 1) + 6] x 7k + 1
= 7 x (6k - 1) x 7k + 6x 7k+1 + 1
= 7 x (6k - 1) x 7k + 7-6 + 42 x 7k
= 7 x [(6k - 1) x 7k + 1] + 6 x (7k - 1)
= 7 x 36N + 6 x (7 - 1) x (7k + 7k-1 + 7k-2 + ... + 1)
= 7 x 36N + 36 x (7k + 7k-1 + 7k-2 + ... + 1)

.'. P(k+1) is divisible by 36 since 7k + 7k-1 + 7k-2 + ... + 1 is a positive integer where k is a positive integer.

By the principle of MI, P(n) is true for all positive integers n.