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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)- i8 N2 [$ [2 G! _$ V& D
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Proof:
8 n- f8 s {9 l1 W+ J% v3 ELet n >1 be an integer 1 K" G3 `7 [$ @% L5 x. [
Basis: (n=2)
% y0 U* G0 w5 O; a* d 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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: ^' A {6 b- O/ \8 z& FInduction Hypothesis: Let K >=2 be integers, support that
A' i7 X1 E- ], | K^3 – K can by divided by 3.8 f4 b9 }; x' N4 E% l$ ?8 D# l0 @
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3- v! R' B: p7 `+ B
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem9 r/ F6 B: _, [' X
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ C7 A9 h' [1 ^7 p$ ^! W, @
= K^3 + 3K^2 + 2K
/ z* a* q+ k% N = ( K^3 – K) + ( 3K^2 + 3K)" M6 B# w4 v1 k2 \
= ( K^3 – K) + 3 ( K^2 + K)1 I! [4 [, b/ \5 I+ s0 k, Z1 g0 n
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
( H1 U5 m; Y4 s: a; J9 kSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)3 [4 }2 |: h" T7 x$ Z4 s$ ^ M
= 3X + 3 ( K^2 + K)
, T* Z0 R3 [. o/ D+ V = 3(X+ K^2 + K) which can be divided by 3
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7 p: W% y1 v* s' fConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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