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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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3 L/ ~- x, V3 R5 t, ?6 v* |Proof: 2 G" T- O3 L) e# p5 A% r1 C' q
Let n >1 be an integer
: b0 u# e% l' i$ w+ tBasis: (n=2)
& S% _" L0 Z6 e/ B1 A( ^0 ? 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3- Q; S! I% d( Z8 y
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Induction Hypothesis: Let K >=2 be integers, support that
" e( ~9 J' g2 N! s/ o, D" E | K^3 – K can by divided by 3.& c% M3 }( i. U/ Z! p# Y& G
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 32 N+ F3 f* b' E% m) k7 n/ B
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
. f5 P2 b" [7 S8 `0 nThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)6 e9 h2 H' D M3 L+ B& P) P/ L+ W% `! v" S
= K^3 + 3K^2 + 2K
/ h. x) ~2 B& ]. U7 E6 _ = ( K^3 – K) + ( 3K^2 + 3K)
& T, Q7 `& g) j, `% h" h7 t% o = ( K^3 – K) + 3 ( K^2 + K)5 \! N0 q- r8 N% e# K
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>05 |3 A4 q4 l* J1 k' H8 h9 E$ A" N
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)& D1 k3 V# I; _& |4 k9 `
= 3X + 3 ( K^2 + K)
. p3 ^* }, n4 C+ H. `: t# G; j = 3(X+ K^2 + K) which can be divided by 34 J% K6 o5 w! y: y
0 j8 Q( b, l+ ^; L6 _" j% BConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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+ u% @4 w# z& c[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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