 鲜花( 53)  鸡蛋( 0)
|
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)
- `# y/ A- A% n% s/ f$ h3 |9 x6 [; k. O7 D7 U
Proof:
/ j4 O- _1 e T4 V* GLet n >1 be an integer 4 K( O9 T8 Z% }( O3 ?& p( t: p
Basis: (n=2)
& L# ]. a% h% |/ J8 \* _6 T' U) J 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3$ G2 \0 k6 D+ H0 ?4 S! S: d$ C6 K: b
* f: ^$ z) d t% c( k; KInduction Hypothesis: Let K >=2 be integers, support that& [) R$ k# t4 s. w! e6 l
K^3 – K can by divided by 3.: R! L8 _8 x( f
# q z3 {# y5 J- f% x
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
& r3 K% ]! Q, [# nsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem" l& A$ a @9 {: M1 ]" w5 ?& H( ]
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)4 w' z, q" j- Q- W# Z
= K^3 + 3K^2 + 2K" s8 D& N3 [. {
= ( K^3 – K) + ( 3K^2 + 3K)
. F& u0 e, I$ G = ( K^3 – K) + 3 ( K^2 + K)/ F) F( @( y5 _" o" R
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 J+ C2 }" f# o3 M: Q
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' _7 U* |; X5 u2 a1 [: L, Z
= 3X + 3 ( K^2 + K)$ e$ C8 A9 M9 P: S
= 3(X+ K^2 + K) which can be divided by 35 ?+ M% T( G) p' ^& M9 Y
- j0 u( U2 k: _8 K" F
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
' \5 T* b4 U. u# b' P7 q6 j- g, d
6 w& p9 |5 _: o3 F[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|