推 GaussQQ :推 04/19 05:24
推 e04su3 :感謝 04/26 22:17
※ 引述《hotplushot (熱加熱)》之銘言：
: 1.
: n為正整數
: (3+根號5)^n + (3-根號5)^n
: 證明其值為2^n的倍數
Denote ω,κ = (3 ±√5)/2,
then
ω,κ satisfy x^2-3x+1=0
ω+κ=3
ωκ= 1
ω^2+κ^2 = (ω+κ)^2 -2ωκ
......
ω^{n+1} + κ^{n+1} = (ω^n + κ^n) (ω+κ) - ωκ(ω^{n-1} + κ^{n-1})
: 3.
: a_n=[1*3*5*...*(2n-1)] / [2*4*6...*2n]
: 求a_n極限 當n趨於無窮大
an = (1/2)*(3/4)*...*(2n-1)/(2n)
bn = (2/3)*(4/5)*...*(2n)/(2n+1)
then an*bn = 1/(2n+1)
and bn > an
hence, 0 < an^2 < an*bn =1/(2n+1)
by squeeze theorem,
an → 0
: 4.
: ax^17+bx^16+1可被x^2-x-1整除
: 求a
: 這四題 煩請高手解題
: 感謝!!!
-x^2-x+1 divides f(x)=-ax^17+bx^16+1
then f(x)=
a_0*(-x^2-x+1)+a_1*(-x^3-x^2+x)+a_2*(-x^4-x^3+x^2)+...
+a_{k-1}*(-x^{k+1}-x^k+x^{k-1})
+a_k*(-x^{k+2}-x^{k+1}+x^k)
+a_{k+1}*(-x^{k+3}-x^{k+2}+x^{k+1})
+...
hence, a_0=1, a_1=1, a_2=2,
a_{k-1} + a_{k} = a_{k+1}, for k=1,...,14
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,
f(x)=987(-x^17-x^16+x^15)+610(-x^16-x^15+x^14)+...
a=987
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◆ From: 123.194.96.239
※ 編輯: JohnMash 來自: 123.194.96.239 (04/19 02:58)