中五(附加數學 MI) 請教

設P(n) 為：對一切正數 n，1/2+ 3/2^2+ 5/2^3+...+2n-1/2^n = 3- (2n+3)/2^n
當n=1, LHS = 1/2; RHS = 3 - (2 + 3)/2 = 1/2 = LHS
假設 P(k)為真，即1/2+ 3/2^2+ 5/2^3+...+ (2k-1)/2^k = 3- (2k+3)/2^k
當n = k + 1 時， 1/2+ 3/2^2+ 5/2^3+...+ (2k-1)/2^k　＋(2k + 1)/2^(k + 1)
= 3- (2k+3)/2^k + (2k + 1)/2^(k + 1)
= 3 - [2(2k + 3) - (2k + 1)]//2^(k +1)
= 3 - (2k + 5)//2^(k +1)
= 3 - [2(k + 1) + 3]//2^(k +1) => P(k + 1)為真
因此以數學歸納法得證

Let S(n) be the statement"1/2+ 3/2^2+ 5/2^3+...+2n-1/2^n = 3- 2n+3/2^n"
when n=1,
LHS=1/2
RHS=3-[2(1)+3]/2^1
=1/2
LHS=RHS
Therefore S(1) is true.
Assume S(k) is true for some positive integer k.
i.e.1/2+ 3/2^2+ 5/2^3+...+2k-1/2^k = 3- 2k+3/2^k

when n=k+1,
LHS=1/2+ 3/2^2+ 5/2^3+...+2k-1/2^k+[2(k+1)-1]/2^(k+1)
=3- 2k+3/2^k+[2(k+1)-1]/2^(k+1)
=3-[2(2k+3)]/2^(k+1)+[2(k+1)-1]/2^(k+1)
=3-[4k+6-2k-1]/2^(k+1)
=3-(2k+5)/2^(k+1)
=3-[2(k+1)+3]/2^k+1
=RHS
S(k+1) is also true.
By the principle of MI,S(n) is true for all positive integers n.