a. maths 問題

As follows~~~


圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/MI20.jpg?t=1190438517

Let P be the proposition "(1-1/4)(1-1/9)(1-1/16)...(1-1/n^2)=(n+1)/2n".
When n=2,
LHS = 1-1/4 = 3/4
RHS = (2+1)/2(2) = 3/4 = LHS
Therefore, P is true for n=2.

Assume P is true for N=k, i.e. (1-1/4)(1-1/9)(1-1/16)...(1-1/k^2)=(k+1)/2k for k≥2.
When n = k+1,
LHS = (1-1/4)(1-1/9)(1-1/16)...(1-1/k^2))[1-1/(k+1)^2]
= [(k+1)/2k][1-1/(k+1)^2]
= [(k+1)/2k] {[(k+1)^2-1]/(k+1)^2}
= [(k+1)^2-1]/2k(k+1)
= [(k+1+1)(k+1-1)]/2k(k+1)
= [k(k+2)]/2k(k+1)
= (k+2)/2(k+1)
= RHS
Hence, P is true for n = k+1.

By the principle of M.I., P is true for all natural numbers n≥2 . (Proved~!!!)

For n=k+1
LHS
= [(k+1)/2k][(1 - 1/(k+1)^2)]

k+1 (k+1)^2 - 1
= (--------)(------------------------)
2k (k+1)^2

(k+1-1)(k+1+1)
= ------------------------------- Be noted that a^2-b^2 = (a+b)(a-b)
2k (k+1)

k * (k+2)
= ---------------
2k(k+1)

= (k+2) / 2(k+1)

= RHS

Therefore by the principles of MI, f(n) is true for all natural numbers n>= 2