F.4 M.I 快~

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Assume S(k) is true.
10^k-3^k = 7M where M is an integer.

when n=k+1
10^k+1-3^k+1 = 10^k*10-3^k+1

= (7M+3^k)*10-3^k+1

= 7M*10+3^k*10-3^k+1

= 7M*10+3^k(10-3)

=7M*10+3^k*7

=7(10M+3^k)

. .
. (10M+3^k) is an integer.
.
. . 10^k+1-3^k+1 is divisible by 7
.
. . S(k+1) is true.

By the principle of mathematical induction, S(n) is true for all positive integer n.

我係讀中中，所以請你將d用語轉番做英文，不過都係咁prove
設P(n)為命題10^n-3^n can divisible by 7
當n=1時
10^1-3^1
=7*1
所以P(1)成立
假設當n=k, p(k)成立
即10^k-3^k=7n(n為一整數)
10^k=7n+3^k
當n=k+1
10^(k+1)-3^(k+1)
=10^k*10-3^k*3
=(7n+3^k)*10-3^k*3
=70n+3^k*10-3^k*3
=70n+(3^k)(10-3)
=70n+(3^k)(7)
=7(10n+3^k)
=7m(m為一整數)
所以根據數學歸納法，對所有正整數n，P(n)都成立