Question

如圖,P1(x1,y1),P2(x2,y2),…,Pn(xn,yn)(0<y1<y2<…<yn)是曲線C:y2=3x(y≥0)上的n個(gè)點(diǎn),點(diǎn)Ai(ai,0)(i=1,2,3,…,n)在x軸的正半軸上,且△Ai-1AiPi是正三角形(A0是坐標(biāo)原點(diǎn)).

 (1) 寫出a1,a2,a3;

(2) 求出點(diǎn)An(an,0)(n∈N+)的橫坐標(biāo)an關(guān)于n的表達(dá)式并用數(shù)學(xué)歸納法證明.

(第5題)

Answer 1

 (1) a1=2,a2=6,a3=12.

(2) 依題意,得xn=,yn=·,

而=3·xn,所以=(an+),即(an-)2=2(+an).

由(1)可猜想:an=n(n+1)(n∈N+).

下面用數(shù)學(xué)歸納法予以證明:

①當(dāng)n=1時(shí),命題顯然成立.

②假設(shè)當(dāng)n=k(k∈N+)時(shí),命題成立,即有ak=k(k+1),

則當(dāng)n=k+1時(shí),由歸納假設(shè)及(-ak)2=2(ak+),

即-2(k2+k+1)+[k(k-1)]·[(k+1)(k+2)]=0,

解得=(k+1)(k+2)或ak+1=k(k-1).因?yàn)?img src='http://thumb.zyjl.cn/pic1/files/down/test/2014/11/13/04/2014111304505093070101.files/image025.gif'>=k(k-1)<ak不合題意,所以舍去,

即當(dāng)n=k+1時(shí),命題成立.

由①,②可知,命題an=n(n+1)(n∈N+)成立.