9^n-2^n能被7整除[[數學歸納法]]

By use of mathematical induction

Is 9^n-2^n divisible by 7

If N = 1
9^1 - 2^1 = 7
therefore N=1 is true

Assume N=k
that is 9^k-2^k=7P for P is any positive integer

If N = k+1
9^(k+1)-2^(k+1)
=9*(9^k)-2*(2^k)
=9*(7P+(2^k))-2*(2^k)
=63P + 9*(2^k) - 2*(2^k)
=63P + 7*(2^k)
=7(9P + (2^k))

Therefore it is true when n=k+1

if n=1 is true, and n=k, n=K+1 is true,
then it must be true for n=1, 2, 3, 4, 5....
ie the statement is true for all integers n > or equal1


Is 6^n-5n+4 divisible by 5

the statement is true when n=1
(working ommited.. too obvious)

let n=k
6^k - 5k + 4 = 5P for any positive integer P

when n=k+1
6^(k+1) - 5(k+1) + 4
= 6*(5P + 5k - 4) - 5(k+1) + 4
= 30P - 25k -15
=5(6P - 5k -3)

ie. statement divisible by 5

S(n)=9^n-2^n
When n = 1
S(1)= 9^1 - 2^1
= 9 - 2
= 7
So s(1)is true.

Assume that S(k) is true for some positive integer.
i.e. 9^k - 2^k = 7M where M is a integer

when n=k+1

S(k+1) = 9^(k+1) - 2^(k+1)
= 9*9^k - 2*2^k
= 9*(7M+2^k) - 2*2^k
=63M+9*2^k-2*2^k
=63M+7*2^k
=7(9M+2^k)
So S(k+1)is true
By the mathematical induction , S(n) is true for all positive integer n.



S(n)=6^n-5^n+4
when n=1
s(1)=6^1-5^1+4
=5
So s(1)is true.

Assume that S(k) is true for some positive integer.
i.e. 6^k-5^k+4 = 5M where M is a integer

when n=k+1

S(k+1) = 6^(k+1)-5^(k+1) +4
= 6*6^k-5*5^k +4
= 6*(5M-4+5^k) -5*5^k +4
= 6*5M-24 +6*5^k - 5*5^k +4
= 6*5M-20+5^k
= 5(6M-4+5^(k-1))
So S(k+1) is true
By the mathematical induction , S(n) is true for all positive integer n.

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9^n-2^n 能被7整除 [[數學歸納法]]

When n = 1, 9^1 - 2^1 = 9 - 2 = 7 能被7整除.

Assume that n = k is true. That is 9^k - 2^k 能被7整除.
Let 7p = 9^k - 2^k
Then
9^(k+1) - 2^(k+1)
= 9*9^k - 9*2^k + 9*2^k - 2^(k+1)
= 9(9^k - 2^k) + 9*2^k - 2*2^k
= 9(7p) + 9*2^k - 2*2^k
= 9(7p) + (9 - 2)*2^k
= 9(7p) + 7*2k
Because 9(7p) 能被7整除 and 7*2k 能被7整除, 9^(k+1) - 2^(k+1)能被7整除.

So 9^n-2^n 能被7整除