 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)
6 b2 s4 u$ q( y, A. u& V# B0 P5 q0 w9 [) G
Proof:
4 T5 G+ E1 ` }. F0 S/ V' F- dLet n >1 be an integer
/ j) W5 R* v& S4 e/ K+ YBasis: (n=2)
: a9 L4 Q x1 T& P 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 31 M, L4 t$ `* A3 x) N
0 E( d5 U8 {7 t3 K, Y6 ]: QInduction Hypothesis: Let K >=2 be integers, support that0 W2 j+ u8 [8 r/ g0 H
K^3 – K can by divided by 3.' q- q1 l: W4 }3 k! k
o- c2 Z$ n! A+ Q+ j5 }
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3% J! z: T/ \- O% C
since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem% x2 M$ f8 d7 ^1 p6 l
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): k5 T7 d# Y- F
= K^3 + 3K^2 + 2K: ^2 f! D9 i4 |/ c1 g9 Z
= ( K^3 – K) + ( 3K^2 + 3K)
- M; R* o, `( V4 J = ( K^3 – K) + 3 ( K^2 + K)
/ x8 w8 u; t: T$ g$ l, Tby 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>01 ~4 y, }6 b& J; o; g0 N& i
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)' P1 \8 j& P3 K) [9 u
= 3X + 3 ( K^2 + K)
, r( V. ^' C. z7 W$ v4 X, C = 3(X+ K^2 + K) which can be divided by 3. n3 x% J+ u5 i2 ~$ `, B
7 W, W' Z" k* T& m! x
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
% T1 H& E$ v5 m' Q7 a0 b& _
6 L# J3 ]2 h: | ~& N, }4 J[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
|