f.4 a. math

When n=1,
LHS=1 - 1/(2*1)=1/2=1/(1+1)=1/2=RHS
So the statement holds for n=1.

Assume the statement holds for n=k, for n=k+1
LHS= 1 - 1/2 + 1/3 + ... + 1/(2k) - 1/(2k+1) + 1/(2k+2)
=1/(k+1) + 1/(k+2) + ... + 1/(2k) + 1/(2k+1) + 1/(2k+2) (By induction hypothesis)
=RHS

So the statement holds for n=k+1 if it holds for n=k
By MI , the statement holds for all positive integer n.

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When n =3, 5^3-3^3-2^3 = 90 = 3*30
So the statement holds for n = 3

Assume the statement is true for n=k, then for n=k+2,
5^(k+2)-3^(k+2)-2^(k+2)
=25*5^k-9*3^k-4*2^k
=36*5^k-20*3^k-15*2^k - 11*(5^k-3^k-2^k)
=36*5^k-20*3^k-15*2^k - 11*30*N (for some integer N, by induction hypothesis)
=6*30*5^(k-1)-2*30*3^(k-1)-30*2^(k-1) - 11*30*N
=30*(6*5^(k-1)-2*3^(k-1)-2^(k-1) - 11*N)
So the statement holds for n=k+2 if it holds for n=k
By MI, the statement holds for all positive odd integers n > 1

2008-11-02 17:02:08 補充：
補充
36*5^k-20*3^k-15*2^k - 11*(5^k-3^k-2^k)
= 36*5^k-20*3^k-15*2^k - 11*5^k+11*3^k+11*2^k
= (36-11)*5^k - (20-11)*3^k - (15-11)*2^k
= 25*5^k-9*3^k-4*2^k

2008-11-02 17:04:34 補充：
補充
36*5^k-20*3^k-15*2^k - 11*30*N
= (6*6)*(5*5^(k-1)) - (2*10)*(3*3^(k-1)) - 15*(2*2^(k-1)) - 11*30*N
= 6*(6*5)*5^(k-1) - 2*(10*3)*3^(k-1) - (15*2)*2^(k-1) - 11*30*N
=6*30*5^(k-1) - 2*30*3^(k-1) - 30*2^(k-1) - 11*30*N