有題M.I.唔識，請幫幫忙

(a)Let P(n) denote 1^3+2^3+3^3+...+n^3=1/4(n^2)(n+1)^2
(1) n=1
LHS=1^3 =1
RHS=1/4 (1)(2)^2 =1
Since LHS = RHS therefore P(1) is true.
(2)Assume p(k) is true ie 1^3+2^3+3^3+...+k^3=1/4(k^2)(k+1)^2
(3) n= k+1
LHS=1^3+2^3+3^3+...+k^3+(k+1)^3
=1/4 (k^2)(k+1)^2+(k+1)^3
=1/4 (k+1)^2(k^2 + 4(k+1))
=1/4 (k+1)^2(k+2)^2
RHS=1/4 (K+1)^2 (k+1+1)^2
=1/4 (k+1)^2 (k+2)^2
LHS=RHS
therefore P(k+1) is true.

By the principle of MI , P(n)is true for all positive integers n.

2009-01-30 00:40:33 補充：
(b)(b) 1^3+3^3+5^3+...+(2m-1)^3
=1^3+2^3+3^3+...+(2m)^3-2^3-4^3-6^3-...-(2m)^3
=1/4 ((2m)^2 (2m+1)^2) -2^3 ( 1^3+2^3+...+m^3)
= 1/4 (2m)^2 (2m+1)^2-8(1/4 (m^2) (m+1)^2)
=m^2 (2m+1)^2 - 2 (m^2) (m+1)^2
=m^2 ((2m+1)^2-2(m+1)^2)
=m^2 (4m^2+4m+1-2m^2-4m-2)
=m^2 (2m^2-1)

(a) Let P(n) be the proposition that

1^3+2^3+3^3+...+n^3=1/4(n^2)(n+1)^2

P(1): L.H.S=1
R.H.S=1/4(1^2)(1+1)^2=1

>P(1) is true.

Assum P(k) is true for some integer k

i.e. 1^3+2^3+3^3+...+k^3=1/4(k^2)(k+1)^2

P(k+1): 1^3+2^3+3^3+...+k^3+(k+1)^3

=1/4(k^2)(k+1)^2+(k+1)^3

=1/4(k^2)(k+1)^2+4/4(k+1)^3 <個4/4係通分母>

=(k+1)^2 [k^2+4(k+1)]
----------------------------
4
=1/4(k+1)^2 (k+2)^2=R.H.S

>P(k+1) is also true.

By Mathematical Induction, P(n) is true for all postion integers n.

(b)Hence,

1^3+3^3+5^3+...+(2m-1)^3

=1^3+2^3+3^3+...+(2m)^3-(2^3+4^3+6^3+...+(2y)^3)
<括孤係計even no.>
=1^3+2^3+3^3+...+(2m)^3-2^3(1^3+2^3+3^3+...+y^3)
<n=2y>(減號既前part) <n=y>(減號既後part)
=1/4(2y^2)(2y+1)^2-2^3[1/4(n^2)(n+1)^2]

=u got the solution now. For any quest, u can contact me
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(b)
1^3+3^3+5^3+...+(2m-1)^3 = 1^3+2^3+3^3+...+(2m)^3 - [2^3+4^3+6^3+...+(2m)^3]
= 1^3+2^3+3^3+...+(2m)^3 - 8(1^3+2^3+3^3+...+m^3)