Mathematical Induction

Let P(n) be the statement "1 + 2(2) + 3(2)^2 + 4(2)^3 + ... + n*2^(n-1) = n*2^(n+1) - (n+1)2^n + 1"

When n=1, L.H.S.=1,R.H.S.=1*4-2*2+1=1; P(1) is true.

Assume that P(k) is true.

When n=k+1

L.H.S.
=1 + 2(2) + 3(2)^2 + 4(2)^3 + ... + k*2^(k-1)+(k+1)*2^(k)
=k*2^(k+1) - (k+1)2^k + 1 + (k+1)*2^(k)
=k*2^(k+1) + 1
=(2k+2-k-2)2^(k+1) + 1
=(k+1)*2^(k+2) - (k+2)2^(k+1) + 1
=R.H.S.

So,P(k+1) is true. By M.I. for all positive n, 1 + 2(2) + 3(2)^2 + 4(2)^3 + ... + n*2^(n-1) = n*2^(n+1) - (n+1)2^n + 1

膠人至上主義者,慢左乜呀???

2010-01-09 20:55:26 補充：
點解??????????

2010-01-10 15:57:34 補充：
膠人至上主義者,你依 家係咪大人?????????????????????????????????

When n=1
LHS = 1*2^0 = 1
RHS =1*2^2 - 2*2^1 +1 =1
so the proposition is true for n=1

assume the proposition is true for some positive integer k
1 + 2(2) + 3(2)^2 + 4(2)^3 + ... + k*2^(k-1) = k*2^(k+1) - (k+1)2^k + 1
Then
1 + 2(2) + 3(2)^2 + 4(2)^3 + ... + k*2^(k-1) + (k+1)*2^k
= k*2^(k+1) - (k+1)2^k + 1+ (k+1)*2^k
= k*2^(k+1) +1
=2k*2^(k+1) - k*2^(k+1) +1
=k*2^(k+2) -k*2^(k+1) +1
=k*2^(k+2) -k*2^(k+1) +1 +2^(k+2) - 2^(k+2)
=(k+1)*2^(k+2) - (k+2)*2^(k+1) +1
hence the proposition is true for n=k+1
By induction, the proposition is true for all positive integer k

2010-01-08 16:46:07 補充：
慢了......真可惜

2010-01-09 14:27:04 補充：
通常數學版答第一個才有最佳

2010-01-10 10:42:30 補充：
數學題有固定答案
當然是最快的有最佳