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This can by done by Induction
Show that for all integers n >1, n^3 - n can be divided by 3. (Note: n^3 stands for n*n*n)
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Proof:
- P7 Q7 ?3 U, n% JLet n >1 be an integer
~' v" w3 R+ D! V+ ]% W3 TBasis: (n=2)
7 {) v' I1 D4 ]* h! H. [ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
t0 y9 |6 y ?2 R- O( r8 ]6 H; R K^3 – K can by divided by 3.% I) q8 E1 K, {; n/ {" }
6 G5 j0 D% x' r; C$ S0 O+ E1 `, mNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 34 O8 |7 _: ]5 Q# A2 ?/ @; b
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem* s2 y% P6 t) k! }
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)* k- ?2 z% W# I) k, [
= K^3 + 3K^2 + 2K( o9 `: K3 w6 V% [# I% c: | Y' R
= ( K^3 – K) + ( 3K^2 + 3K)
$ N6 e# q3 T" {1 k4 R = ( K^3 – K) + 3 ( K^2 + K)7 ?! z) h* U: l( [" Q
by Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>08 Z9 O5 D f0 g: t0 R
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
+ b f- n0 K; w- E = 3X + 3 ( K^2 + K)5 N3 C4 v8 | M& Z& ^0 n
= 3(X+ K^2 + K) which can be divided by 30 |3 M+ C$ Y; u) ^, k+ \
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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5 j% V `$ A8 ~+ G; A' b[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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