中四A-MATHS題- o -

我老師就係咁教我既,唔知你係唔係 " ^ " = 次方
有d格式唔記得,所以最後個d因為∴可能唔岩,應該計得岩,
2^3 4^3 6^3 ... (2n)^3=2n^2(n 1)^2
解:
設S(n)為命題:2^3 4^3 6^3 ... (2n)^3=2n^2(n 1)^2
當n=1時:
左方:[2(1)]^3=8
右方:2(1)^2(1 1)^2=2(2)^2=8
因為左方=右方
∴n=1成立,即S(1)成立
當n=k:
即:2^3 4^3 6^3 ... (2k)^3=2k^2(k 1)^2
當n=k 1:
左方=2^3 4^3 6^3 ... [2(k 1)]^3
=2k^2(k 1)^2 [2(k 1)]^3
=2k^2(k 1)^2 8(k 1)^3
=2(k 1)^2[k^2 4(k 1)]
=2(k 1)^2(k^2 4k 4)
=2(k 1)^2(k 2)^2
=右方
∴n=k 1成立,S(k 1)成立
根據數學.......(太長不寫)

when n=1
LHS=(2n)^3 =8
RHS= 2n^2(n+1)^2=2 x 4 =8
n=1 is ture
Assume n=k is true
ie: 2^3 + 4^3 + 6^3 +...+(2k)^3=2k^2(k+1)^2
Consider n=k+1
LHS= 2^3 + 4^3 + 6^3 +...+(2k)^3 + (2k+2)^3
= 2k^2(k+1)^2 +(2k+2)^3
=2k^2(k+1)^2 +8(k+1)^3
=2(k+1)^3[k^2 + 4k +4] ←e度係抽左2(k+1)^3 出黎
=2(k+1)^3(k+1+1)^2
=RHS
By the prove of MI, it is true of all positive integers of n
我唔知中文點寫,我諗你對番寫textbook就ok

2^3 + 4^3 + 6^3 +.......+ (2n)^3

= 2^3 ( 1^3 +2^3 +3^3 +..... n^3)

= 8 x [n^2(n+1)^2]/4

= 2n^2(n+1)^2



( 1^3 +2^3 +3^3 +..... n^3) = [n^2(n+1)^2]/4 公式黎,好難解(超長)

2008-10-13 22:37:26 補充：
( 1^3 +2^3 +3^3 +..... n^3) = [n^2(n+1)^2]/4
詳情查下咩係sumation就得