F.4maths

f(1):

LHS = 1*2^0 = 1*1 = 1
RHS 1+(0)2^1 = 1 = LHS
therefore, f(1) is true

Assume f(n)is true:
then 1*2^0+2*2+3*2^2+...+n*2^(n-1)=1+(n-1)2^n is true for all +ve integer n

for f(n+1):

LHS:
= 1*2^0+2*2+3*2^2+...+ n * 2^(n - 1) + (n+1) * 2^n
=f(n) + (n+1)*2^n
= 1 + (n - 1) 2^n +(n+1)*2^n
= 1 +2n * (2^n)
= 1+ n * 2^(n + 1)

RHS:
=1 + (n + 1 - 1) 2^(n + 1)
= 1+ n* 2^(n + 1) = LHS
f(n+1) is true

Therefore by the mathematical induction f(n) is true for all positive integrs n. #



2. A(n)=1-(2/2n+1) = (2n - 1)/(2n+1) and T(n)=A(1)*A(2)*A(3)*....A(n)

T(n) = [1/3] * [3/5] * [5/7] * ......* [(2n - 3)/(2n - 1)] * [(2n - 1)/(2n + 1)]

T(n) = 1 / (2n + 1) #

2010-10-11 04:03:09 補充：
T(n) =[1/3] * [3/5] * [5/7] * ......* [(2n - 1)/(2n + 1)] = 1 / (2n + 1)

T(1):
LHS = 1/3
RHS = 1/3 = LHS
therefore T(1) is true

2010-10-11 04:04:19 補充：
Assume T(n) is true, then
T(n) =[1/3] * [3/5] * [5/7] * ......* [(2n - 1)/(2n + 1)] = 1 / (2n + 1)
is true

2010-10-11 04:04:30 補充：
For T(n+1)
LHS
= [1/3] * [3/5] * [5/7] * ......* [(2n - 1)/(2n + 1)] * [(2{n + 1} - 1)/(2{n + 1} + 1)]
= [1/3] * [3/5] * [5/7] * ......* [(2n - 1)/(2n + 1)] * (2n + 1)/(2n + 3)
= T(n) * (2n + 1)/(2n + 3)
= [1 / (2n + 1)] * [(2n + 1)/(2n + 3)]
= 1/(2n+3)

2010-10-11 04:04:38 補充：
RHS
= 1(2n+3) = LHS

therefore T(n+1) is true

Therefore by the mathematical induction T(n) is true for all positive integrs n. #