云南昭通實(shí)驗(yàn)中學(xué)高中數(shù)學(xué)《等比數(shù)列前n項(xiàng)和》課件9 新人教A必修5

等比數(shù)列的前n項(xiàng)和,等差數(shù)列an,等比數(shù)列an,定義,an+1-an=d(常數(shù)),an+1an=q(不為零的常數(shù)),通項(xiàng),an=a1+(n1)d,an-am=(nm)d,an=a1qn-1,anam=qn-m,公式,推導(dǎo)方法,歸納猜想驗(yàn)證法,首尾相咬累加法,歸納猜想驗(yàn)證法,首尾相咬累乘法,性質(zhì),若m+n=r+s,m、n、r、sN*,則am+an=ar+as,若m+n=r+s,m、n、r、sN*,則aman=aras,前n項(xiàng)和Sn,公式,推導(dǎo)方法,化零為整法,問(wèn)題：等比數(shù)列an，如果已知a1,q,n怎樣表示Sn?,Sn=a1+a2+an,解：,=a1+a1q+a1q2+a1qn-1,=a1(1+q+q2+qn-1),嘗試：,S1=a1,S2=a1+a1q=a1(1+q),S3=a1+a1q+a1q2=a1(1+q+q2),討論q1時(shí),猜想：,Sn,驗(yàn)證：,an=Sn-Sn-1,=a1qn-1,當(dāng)n2時(shí),an=a1qn-1,證明（）式,(1+q+q2+qn-1)(1-q),=1+q+q2+qn-1,-(q+q2+qn-1+qn),=1-qn,（）式成立,相減,(1q)Sn=a1-a1qn,=a1(1qn),當(dāng)1q0,即q1時(shí)，,當(dāng)q=1時(shí)，,Sn=na1,錯(cuò)項(xiàng)相減法：,Sn=a1+a1q+a1q2+a1qn-1,qSn=a1q+a1q2+a1qn-1+a1qn,等比數(shù)列an前n項(xiàng)和公式為,當(dāng)q1時(shí),當(dāng)q1時(shí),Sn=na1,264-1,-4或3,例1：求通項(xiàng)為an=2n+2n-1的數(shù)列的前n項(xiàng)和,解：,=2(2n1),=n2,解：,當(dāng)x=1時(shí),Sn=,當(dāng)x1時(shí),Sn=,n+,(2)只須注意再討論y是否等于1的取值情況,例3：求數(shù)列：1,2x,3x2,，nxn-1,（x0)的前n項(xiàng)和,解：,當(dāng)x=1時(shí)Sn=1+2+3+n=,當(dāng)x1時(shí)Sn=1+2x+3x2+nxn-1,xSn=x+2x2+(n-1)xn-1+nxn,錯(cuò)項(xiàng)相減,(1x)Sn=1+x+x2+xn-1-nxn,-nxn,等差數(shù)列an,等比數(shù)列an,定義,an+1-an=d(常數(shù)),an+1an=q(不為零的常數(shù)),通項(xiàng),an=a1+(n1)d,an-am=(nm)d,an=a1qn-1,anam=qn-m,公式,推導(dǎo)方法,歸納猜想驗(yàn)證法,首尾相咬累加法,歸納猜想驗(yàn)證法,首尾相咬累乘法,性質(zhì),若m+n=r+s,m、n、r、sN*,則am+an=ar+as,若m+n=r+s,m、n、r、sN*,則aman=aras,前n項(xiàng)和Sn,公式,推導(dǎo)方法,=na1+,化零為整法,方法三：,Sn=a1+a2+an,=a1+a1q+a1q2+a1qn-1,=a1+q(a1+a1q+a1qn-2),=a1+qSn-1,=a1+q(Snan