A.maths 高手請答

5^2x 4^(y+1) =25
log[5^2x 4^(y+1)]=log25
log 5^2x+log4^(y+1)=log5^2
2xlog5+2(y+1)log2=2log5
xlog5+(y+1)log2=log5 --------------(1)

4^x 5^2y =1
log[4^x 5^2y]=log 1
log4^x+log5^2y=0
2xlog2+2ylog5=0
x=-(ylog5/log2) -------------(2)

sub (2) into (1)
-(ylog5/log2)log5+(y+1)log2=log5
y[(log 2}^2 - (log 5)^2]=log2(log5-log2)
y(log 2+ log 5)(log2 - log5)=log2(log5-log2)
y=-log2/(log5+log2)=-log2/log(5*2)=-log2/log10=-log2

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Let Q(n):1+√5)^n + (1 -√5)^n is divisible by 2^n

For n=1,
(1+√5) + (1 -√5)=2 is divisible by 2

Assume Q(k) is true
(1+√5)^k + (1 -√5)^k =(2^k)m for m is integer

For n = k+1,
諗諗先..

2007-06-15 16:16:37 補充：
eiffeltam的方法真妙...不過我想補充一點...就是證明 S(k) = 2S(k-1) 4S(k-2)

2007-06-15 16:16:51 補充：
2S(k-1) 4S(k-2)=2[(1 √5)^(k-1) (1 -√5)^(k-1)] 4[(1 √5)^(k-2) (1 -√5)^(k-2)]=(1 √5)^(k-2)[2(1 √5) 4] (1 -√5)^(k-2)[ 2(1 -√5) 2]=(1 √5)^(k-2)[1 2√5) 5] (1 -√5)^(k-2)[ 1 -2√5 5]=(1 √5)^(k-2)(1 √5)^2 (1 -√5)^(k-2)(1 -√5)^2=(1 √5)^k (1 -√5)^k=S(k)

Q2.

假設 S_n = (1+sqrt{5})^n + (1-sqrt{5})^n。

我們可以先用簡單的運算證明 S_{n+2}=2S_{n+1}+4S_n。

之後，先證明頭兩個cases(即n=1,2)命題成立，再假設n=k,k+1時命題成立，就可證明出n=k+2時命題成立。在此我只給出induction step的步驟。

Assume S_k and S_{k+1} is divisible by 2^k and 2^{k+1} respectively, i.e. there exist two integers P and Q such that S_k = 2^k P and S_{k+1} = 2^{k+1} Q.

For n=k+2,
S_{k+2}
= 2S_{k+1} + 4S_k
= 2(2^{k+1} Q) + 4(2^k P)
= 2^{k+2} Q + 2^{k+2} P
= 2^{k+2} (P+Q).

Since P+Q is an integer, S_{k+2} is divisible by 2^{k+2}.