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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): ]; x |4 |$ n- P' N) y# S
. ^ \4 Z5 U3 QProof: / l# e/ m( o/ @
Let n >1 be an integer
/ ^. i+ i+ |$ B5 A8 vBasis: (n=2)3 w9 i( K+ ~8 y% F8 t, c- P
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3; x8 `/ a" f- B1 Q) j
) L* J2 C2 A& l! A1 Q# o; I" b bInduction Hypothesis: Let K >=2 be integers, support that
9 P! p. O5 P) M9 D5 n0 d% h! j$ R K^3 – K can by divided by 3.
1 Q {3 o5 T5 b4 M+ d$ D# f- q
# q' `1 R7 C5 \6 [0 ^' a& \Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
. t: q. f5 T: Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem( I9 w3 P# W3 L u8 j9 x
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
& ?- R" X6 e* W = K^3 + 3K^2 + 2K( k' I' |9 P2 A% L n) @8 m
= ( K^3 – K) + ( 3K^2 + 3K) Y' F0 z6 E; Z" D5 F
= ( K^3 – K) + 3 ( K^2 + K)
D2 Y' D/ b0 H5 ?+ ^" ~: m6 @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 o7 ~: M9 Z7 o5 X1 Q. b
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
% H4 E2 T. D; h2 n) t& c = 3X + 3 ( K^2 + K)
- D9 t3 {+ P: b& S" m4 S = 3(X+ K^2 + K) which can be divided by 38 w, B. o4 W. N% P7 n2 U* E# i
$ z. k7 H# c; B2 E: \Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
0 B( @. P j+ N% Z! a2 e* V+ ~2 i% s9 x: X
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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