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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)4 F& M" \3 I9 Q! v4 { P1 {" n
5 G; R6 G \* w: i; z" G4 o' YProof:
1 v' L9 b+ x; c# q5 f) O" g5 mLet n >1 be an integer
6 P G# t" m: X1 g- Y. l2 @ NBasis: (n=2)+ N! }2 J4 U+ \1 z* f3 h
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3. S. g, p ~: O0 r( r
1 C( y, v+ l6 T5 QInduction Hypothesis: Let K >=2 be integers, support that M* y) d* Y1 j
K^3 – K can by divided by 3.. x' e9 P/ e" M) ?4 G, d7 ~
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
) K8 Y4 p' i; r, M! y; s7 ysince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem1 i9 j5 _+ X7 J: i2 u ^
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
4 }7 r, r7 ?/ M6 P/ n = K^3 + 3K^2 + 2K
' M$ k8 C5 p. t0 m = ( K^3 – K) + ( 3K^2 + 3K)+ C8 X. }6 w. [4 V! d0 F% S5 S% y
= ( K^3 – K) + 3 ( K^2 + K): `& ~3 Z4 H2 L# Y( ]* T! w' ~. B, U7 L
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
9 m$ u/ k0 Q4 S8 B, RSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
5 o& O. u/ d# ^5 \3 j8 [ = 3X + 3 ( K^2 + K)4 u& f" W! d2 ]" V8 t' w
= 3(X+ K^2 + K) which can be divided by 3
" e4 P6 o$ V6 B8 B/ ]* h9 Q5 n6 U& x; I9 l
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1., J% v: L2 H5 V$ d2 x8 ]$ l
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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