F.4 A.Math - Mathematical Induction

2006-11-02 3:40 am
(a) If a=/= +1 or -1 ,prove by mathematical induction that

 1    2    4       2^n
--------- + ---------- + ---------- + .... + ---------- =
1 + a  1+a^2  1+a^4     1+a^2^n
 1    2^(n+1)
--------- + ------------------ for all positive integer n.
a - 1  1-a^2^(n+1)


(b) If a=1 ,write down a formula for
 1    2    4      2^n
--------- + ---------- + ---------- + .... + ---------- for all positive integer n.
1 + a  1+a^2  1+a^4     1+a^2^n

回答 (1)

2006-11-02 10:17 am
✔ 最佳答案
(a) Let P(n)= 1    2    4       2^n
--------- + ---------- + ---------- + .... + ---------- =
1 + a  1+a^2  1+a^4     1+a^2^n
 1    2^(n+1)
--------- + ------------------ for all positive integer n.
a - 1  1-a^2^(n+1)

when n=1,
L.H.S. = 1/(1+a)+2/(1+a^2)
= [(1+a^2)+2(1+a)]/(1+a)(1+a^2) (通分母)
= (a^2+2a+3)/(1+a)(1+a^2)

R.H.S. = 1/(a-1) + 2^(1+1)/[1-a^2^(1+1)]
= 1/(a-1) + 4/(1-a^4)
= 1/(a-1) + 4/(1-a^2)(1+a^2)
= 1/(a-1) + 4/(1-a)(1+a)(1+a^2)
= 1/(a-1) - 4/(a-1)(1+a)(1+a^2)
= [(a+1)(a^2+1)-4]/(a-1)(1+a)(1+a^2) (通分母)
= (a^3+a^2+a-3)/(a-1)(1+a)(1+a^2)
做到呢度,發覺左邊同右邊既分母差左a-1
所以可以試下用長除將右邊既分子除a-1
so, = (a-1)(a^2+2a+3)/(a-1)(1+a)(1+a^2) (省略左長除既步驟,希望你會明)
= (a^2+2a+3)/(1+a)(1+a^2)
= L.H.S.
So, k=1 is true
Assume P(k) is true where k is a positive integer
i.e.
1   2    4       2^k
--------- + ---------- + ---------- + .... + ---------- =
1 + a  1+a^2  1+a^4     1+a^2^k
 1    2^(k+1)
--------- + ------------------ for all positive integer k.
a - 1  1-a^2^(k+1)

Want to prove P(k+1) is true.

When n=k+1,

  1    2    4       2^k 2^(k+1)
--------- + ---------- + ---------- + .... + ---------- + ---------------------
1 + a  1+a^2  1+a^4     1+a^2^k 1+a^2^(k+1)

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

= 1/(a-1) + 2^(k+1) [1+a^2^(k+1)+1-a^2^(k+1)]/[1-a^2^(k+1)][1+a^2^(k+1)] (通分母)

= 1/(a-1) + 2^(k+1)(2)/[1-a^2^(k+1) a^2^(k+1)]

= 1/(a-1) + 2^[(k+1)+1]/[1-a^2^(k+1)(2)]

= 1/(a-1) + 2^[(k+1)+1]/[1-a^2^(k+1)(2^1)]

= 1/(a-1) + 2^[(k+1)+1]/{1-a^2^[(k+1)+1]}
So, P(k+1) is true for k is a positive integer

By the principle of M.I., P(n) is true for all positive integer n

(b) 1    2    4      2^n
--------- + ---------- + ---------- + .... + ---------- for all positive integer n.
1 + a  1+a^2  1+a^4     1+a^2^n

If a=1,

1    2    4      2^n
--------- + ---------- + ---------- + .... + ---------- for all positive integer n.
1 + 1  1+1^2  1+1^4     1+1^2^n


=1/2+2/2+4/2+...+2^n/2

=1/2 (1+2+4+...+2^n)

=(1/2) [2^(n+1)-1]/(2-1) (by using G.P. formula, first term is 1 and common ratio is 2)

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

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

(c) For a=3,

P(99)= 1/4+2/10+4/82+...+ 2^99/(1+3^2^99)

So, 2P(99) = 1/2+2/5+4/41+...+2^100/(1+3^2^99)
= 2 {[1/(3-1)]+2^(99+1)}/{1-[3^2^(99+1)]} by part a
= 2{[1/2]+2^100}/{1-[3^2^100]}
= 1+{2^101/[1-(3^2^100)]}
參考: 自己...都用左幾耐時間


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