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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)
+ a) p% {% a8 h+ H/ q( Q/ q4 m5 I% ?9 E
Proof: % ^- g9 v) J, x6 Y
Let n >1 be an integer
& B' i5 ?/ M3 a* d' Z1 @0 rBasis: (n=2)* b) R: z- Q: r d" H; b( ~+ B: q! V
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3: d* W0 V. m: B2 e% f
' G% C4 }+ k+ E6 P8 MInduction Hypothesis: Let K >=2 be integers, support that
2 i' Q6 E8 _8 N& [7 f K^3 – K can by divided by 3.
1 `% s4 ^+ B1 j( }8 M4 E+ M0 ?- Z' `9 D# K' z& J4 l
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3& d" E: B; b# Q8 Z7 G
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem4 F) G8 w1 R3 i
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)% [# V. C; t/ _) M* Z
= K^3 + 3K^2 + 2K: C6 J* C: F+ {. h- n5 c5 `7 z
= ( K^3 – K) + ( 3K^2 + 3K)
, L6 z+ I u5 m X& r$ e = ( K^3 – K) + 3 ( K^2 + K)6 e* e$ O! s# n; P0 I. A. c
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 B$ t0 h/ j, j5 ?
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)+ X4 V" C8 L9 Y- P! ]) n# k0 m
= 3X + 3 ( K^2 + K)
5 l0 [0 k* X( x Y = 3(X+ K^2 + K) which can be divided by 3" E3 E$ J2 o: w0 G+ @
4 o$ [( h8 h I& u) Q% R$ TConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.2 Z8 P8 X% H. A
$ Q/ j' v: d8 ~! g' Y1 T+ {: X[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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