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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)$ U2 E: k3 e. G* P" \- X L9 V
# W3 {; z0 A1 I" |* q9 mProof: : h3 R4 H2 W5 M3 X
Let n >1 be an integer
' o _& z+ m2 h4 [" N) c) W. M; eBasis: (n=2)
9 A3 A+ A+ v J$ R6 F$ V 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 39 z1 L. c4 l5 G3 M# e2 d2 A
3 L3 w5 [+ Y( z: l
Induction Hypothesis: Let K >=2 be integers, support that
- q! D$ G( a4 y4 x" v' O K^3 – K can by divided by 3.2 P9 Q3 l a1 S }: V5 q& \6 Z u
5 Z5 X+ z; R! r* x4 ^) ], W0 nNow, we need to show that ( K+1)^3 - ( K+1) can be divided by 3/ u* G% Q1 H) s8 [
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem: }+ f" j3 ~" f) x x
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)' C* V4 |; B& w6 ]% m) {' m
= K^3 + 3K^2 + 2K8 R) C2 O0 X) X/ z! B; m
= ( K^3 – K) + ( 3K^2 + 3K)
7 L0 t6 d3 | `0 v+ B' k& B& t = ( K^3 – K) + 3 ( K^2 + K)1 L% I- }" J6 g& r, p2 O
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>0, X6 ~# r O' E4 j2 O) |
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)* O, a9 ?- t5 p# v0 J
= 3X + 3 ( K^2 + K)
# F% k; t. H8 P6 C6 J = 3(X+ K^2 + K) which can be divided by 3* Y% i3 d, k! B
% H) t S9 x/ K6 k7 h* J
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.5 `' u1 w0 m" \" H" T/ M2 Q" [
1 g" x0 Z* q9 C( K) C3 Z! m[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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