Mathematical Induction

there are three step in MI

step 1 : prove that is true for n=1

step 2 : assume it is true for n=k , k is any positive integer

step 3 : prove that is true for n=k+1 ( you may use the assumption in step 2 )

Solution :

when n=1 ( step1 )

LHS = 1^3 = 1
RHS = 1^2(2*1+1)^2/9=1=LHS

so it is true for n=1

Assume it is true for n =k , where k is a positive integer ( step 2 )
i.e. 1^3+2^3+...+k^3=(1/9)k^2(2k+1)^2

Consider n=k+1 (step 3)

1^3+2^3+3^3+....+k^3+(k+1)^3
=(1/9)k^2(2k+1)^2+(k+1)^3 ( by assumption )
=(1/9)k^2(2k+1)^2+(1/9)9(k+1)^3
=(1/9)(4k^4+4k^3+k^2+9k^3+27k^2+27k+9)
=(1/9)(4k^4+13k^3+28k^2+27k+9)
=(1/9)(k+1)^2(2k+3)^2
=(1/9)(k+1)^2(2(k+1)+1)^2

so it is true for n=k+1

By Mathematical induction , it is true for all positive integer n .

2010-02-28 21:23:05 補充：
sorry , i do something wrong =]

2010-02-28 21:23:49 補充：
when n=1 ( step1 )

LHS = 1^3 +2^3= 9
RHS = 1^2(2*1+1)^2=9=LHS

so it is true for n=1

2010-02-28 21:24:08 補充：
Assume it is true for n =k , where k is a positive integer ( step 2 )
i.e. 1^3+2^3+...+k^3+...+(2k)^3=k^2(2k+1)^2

2010-02-28 21:28:24 補充：
Consider n=k+1 (step 3)
1^3+2^3+3^3+....+k^3+...+(2k)^3+...+(2k+2)^3
=1^3+2^3+...+(2k)^3+(2k+1)^3+(2k+2)^3
=k^2(2k+1)^2+(2k+1)^3+(2k+2)^3 ( by assumption )
=(2k+1)^2(k^2+2k+1)+(2k+2)^3

2010-02-28 21:28:28 補充：
=(2k+1)^2(k+1)^2+8(k+1)^3
=(k+1)^2((2k+1)^2+8(k+1))
=(k+1)^2(4k^2+4k+1+8k+8)
=(k+1)^2(4k^2+12k+9)
=(k+1)^2(2k+3)^2
=(k+1)^2(2(k+1)+1)^2

2010-02-28 21:29:00 補充：
so it is true for n=k+1

By Mathematical induction , it is true for all positive integer n