F4 maths ( MI ) 廿分哦

a)
when n=1 ,
1/4(1)^2(1+1)^2=1
L.H.S.=R.H.S.
Assume for some positive integer k
i.e. 1^3+2^3+3^3+...+k^3=1/4k^2(k+1)^2
For n=k+1
1^3+2^3+3^3+...+k^3+(k+1)^3
=1/4k^2(k+1)^2+(k+1)^3 (by assumption)
=1/4[k^2(k+1)^2+4(k+1)^3]
=1/4[(k+1)^2*[k^2+4(k+1)]]
=1/4[(k+1)^2*(k^2+4k+4)]
=1/4(k+1)^2(k+2)^2
By M.I. , the statement is true for all positive integers n
b)
2^3+4^3+6^3+...+(2n)^3
=2^3+(2*2)^3+(2*3)^3+...+(2*n)^3
=2^3+2^3*2^3+2^3*3^3+...2^3*n^3
=2^3(1^3+2^3+3^3+...+n^3)
=8[1/4n^2(n+1)^2] (by a)
=2n^2(n+1)^2
c)
1^3+3^3+5^3+...+(2n+1)^3
=[1^3+2^3+3^3+...(2n+1)^3]-[2^3+4^3+6^3+...+(2n)^3]
=1/4(2n+1)^2(2n+2)^2-2n^2(n+1)^2
=2n^4+8n^3+11n^2+6n+1

2009-07-30 23:51:44 補充：
其實A part的MI係設立左個statement :
Let the statement P(n) be "1^3+2^3+3^3+...+n^3=1/4n^2(n+1)^2"for all positive integers n
而完成這個MI主要分3個部份
1: 證明命題P(1)成立 , 即P(n)命題於n=1時正確
2: 假設有一正整數k令命題正確
3: 再證明P(n)於n=k+1時也正確,根據induction,就會完成這個MI了

至於b,c part係基本的延伸題,記住這2part的技巧便可以!

2009-07-31 00:44:06 補充：
因為(a)-(b)會刪除左個d雙數項,如2^3,4^3,..呢d
做成得番單數項,姐係c part的模樣~