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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)$ L5 V- T! S9 F* h+ q$ A
3 W1 q0 G: E. |2 u) TProof:
9 d. s; ~1 Y p6 g; y- D8 S2 BLet n >1 be an integer ( S/ w( i. l# G
Basis: (n=2)
. O! v! B: r: f4 e 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
: {) E# T2 }8 B. p! l% A
; }/ M2 O7 n$ X6 Y4 ~Induction Hypothesis: Let K >=2 be integers, support that
* N$ w0 P$ ?: ~9 M. h* H9 ?- \% D K^3 – K can by divided by 3.
( _& {, r0 j( W$ e1 u/ J, n0 D; p, G( p+ L# b5 n
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 35 t# k c, d7 h# c0 {
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem, x$ M7 p+ p( Q T8 ^1 B% `9 t$ b
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
) I( T7 Q# t2 n q9 y \# F = K^3 + 3K^2 + 2K R8 F9 \3 \# A$ v$ p" w
= ( K^3 – K) + ( 3K^2 + 3K)2 v& X8 {5 I6 V8 x
= ( K^3 – K) + 3 ( K^2 + K)! q* B5 z) k0 |4 U
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" V2 H& d6 |7 a% X h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
/ U$ O- n# V0 A- F5 V = 3X + 3 ( K^2 + K)
1 r6 r/ z. t1 J' F# Q1 T, g = 3(X+ K^2 + K) which can be divided by 3- {0 x% M3 a: b. s5 q) h4 h& J
) `$ F: D4 p4 C4 I2 }4 W
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1./ {3 ]8 l6 m6 A
+ ~$ \# [, U8 d5 u7 H
[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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