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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)) u+ A( m; f4 z! f" q2 a, {/ n# U
- T' E% X3 L3 c5 u# Q- ]" x( gProof:
& }' ]$ S5 k5 F' C) m. X" N6 @6 nLet n >1 be an integer 5 s1 b1 ]& u) ]: r
Basis: (n=2): E# h* S8 s1 T; r
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
3 p) k4 E% X- W" y) m' V( F$ i* x
Induction Hypothesis: Let K >=2 be integers, support that3 t( F Z1 |" ] L$ w L8 i
K^3 – K can by divided by 3.
# s3 e' k2 Y s5 M" X6 }3 G: n2 u+ `' Z# n0 r
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
' D2 X' T4 _$ C! S* u; Asince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem8 |8 H& f" ^! N8 m0 d# u
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)1 X/ p& o3 c6 L& W4 A0 r
= K^3 + 3K^2 + 2K
: w' o* A, H- |1 B/ k, k& l+ | = ( K^3 – K) + ( 3K^2 + 3K). T" @5 y( U# @. u4 C
= ( K^3 – K) + 3 ( K^2 + K)5 T/ i( d$ L- ]3 n. y% \
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>00 b" S P6 |+ K! q1 B/ \! G
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
2 V" T; d, b. [- W: P = 3X + 3 ( K^2 + K)
6 H. W, E$ _, h/ y7 ? = 3(X+ K^2 + K) which can be divided by 3
& `# O# I8 w- R0 y0 V. _2 R
: C0 }9 ~4 ]. n `Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.( i" `9 H8 B0 D2 ~( i2 @+ v& G
9 y' x0 w/ ~6 D! |5 I& F! p6 W[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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