✔ 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)]}