Mathematical Induction

2007-10-10 4:21 am
Prove that 5^n-3^n-2^n isdivisible by 30 for all positive odd integers greater than 1.

回答 (2)

2007-10-10 4:49 am
✔ 最佳答案
Prove that 5^n-3^n-2^n is divisible by 30 for all positive odd integers greater than 1
5^3 - 3^3 - 2^3
= 90
is divisible by 30
assume 5^k - 3^k - 2^k is divisible by 30, k is odd integer greater than 1
5^k - 3^k - 2^k = 30m, where m is integer
consider
5^(k+2) - 3^(k+2) - 2^(k+2)
= (25)5^k - (9)3^k - (4)2^k
= (25) [ 30m + 3^k + 2^k ] - (9)3^k - (4)2^k
= (25)(30m) + (25)3^k + (25)2^k - (9)3^k - (4)2^k
= (25)(30m) + (16)3^k + (21)2^k
去到呢步, 停一停, 去證明下面另一條 M.I. 先
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(16)3^k + (21)2^k is divisible by 30 for k odd integer greater than 1
(16)3^3 + (21)2^3 = 600 is divisible by 30
assume (16)3^p + (21)2^p is divisible by 30, p is odd integer greater than 1
(16)3^p + (21)2^p = 30q, q is integer
consider
(16)3^(p+2) + (21)2^(p+2)
= (9)(16)3^p + (84)2^p
= (9) [ 30q - (21)2^p ] + (84)2^p
= (9) 30q - (189)2^p + (84)2^p
= (9) 30q - (105)2^p
= (9) 30q - (210)2^(p-1)
= (30) [ 9q - (7)2^(p-1) ]
because 9q - (7)2^(p-1) is integer
(16)3^(p+2) + (21)2^(p+2) is divisible by 30
so, when case k = p is true, case k = p+2 is also true
by M.I., (16)3^k + (21)2^k is divisible by 30 for k odd integer greater than 1
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好, 返回第一條 M.I.
(25)(30m) + (16)3^k + (21)2^k 證明到
is divisible by 30
所以第一條都證明完,
明未?
2007-10-10 9:51 am
The statement is " 5^n-3^n-2^n is divisible by 30 for all positive odd integers greater than 1 "

Step 1: Prove the statement is true for n=3
When n=3, 5^3-3^3-2^3 = 90 = 3(30) which is divisble by 30

Step 2: Assume the statement is also true for n=k (where k is an odd integer greater than 1)
When n=k, 5^k - 3^k - 2^k = 30M ( where M is an integer )

Step 3: Prove the statement is true for n=k+2
When n=k+2,
5^(k+2)-3^(k+2)-2^(k+2)
= 25(5^k) - 9(3^k) - 4(2^k)
= 25(30M + 3^k + 2^k) - 9(3^k) - 4(2^k)
= 25(30M) + (25-9)(3^k) + (25-4)(2^k)
= 25(30M) + 16(3^k) + 21(2^k)

Repeat Step 1-3 to prove the statement " 16(3^n) + 21(2^n) is divisible by 30 for all positive odd integers greater than 1 " is true.
You should be able to prove that 16(3^k) + 21(2^k) = 30P (where P is an integer)

= 25(30M) + 16(3^k) + 21(2^k)
= 30(25M + P)
which is divisible by 30.


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