高中數(shù)學(xué)《等比數(shù)列前n項(xiàng)和》9 新人教A必修5

等比數(shù)列的前n項(xiàng)和精選ppt等差數(shù)列{an}等比數(shù)列{an}定義an+1-an=d(常數(shù)

)an+1／

an=q(不為零的常數(shù)

)通項(xiàng)an=a1+(n–1)dan-am=(n–m)dan=a1qn-1an／

am=qn-m⑴公式⑵推導(dǎo)方法①歸納猜想驗(yàn)證法②首尾相咬累加法①歸納猜想驗(yàn)證法②首尾相咬累乘法性質(zhì)若m+n=r+s,m、n、r、s∈N*則am+an=ar+as若m+n=r+s,m、n、r、s∈N*則am·an=ar·as前n項(xiàng)和Sn⑴公式⑵推導(dǎo)方法（a1+an）nSn=

2=na1+n(n–1)

2d化零為整法精選ppt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)精選ppt嘗試：S1

=a1S2

=a1+a1q=a1(1+q)S3

=a1+a1q+a1q2=a1(1+q+q2)討論q≠1時(shí)

a1(1–q3

)1-q=

a1(1–q2

)1-q=

a1(1–q1

)1-q=猜想：Sn

a1(1–qn

)1-q=驗(yàn)證：an=Sn-Sn-1

a1(1–qn)1-q=-

a1(1–qn-1)1-q=a1qn-1

a1(qn-1–qn)1-q=當(dāng)n≥2時(shí)當(dāng)n=1時(shí)a1=S1

亦滿足上式∴an=a1qn-1∴Sn(q≠1)

a1(1–qn)1-q=精選ppt

a1(1–qn)1-q=Sn=a1+a2+···+an=a1+a1q

+a1q2

+···+a1qn-1=a1(1+q+q2+···+qn-1)當(dāng)q≠1時(shí)即1+q+q2+···+qn-1…………（＊）

1–qn1-q=證明（＊）式(1+q+q2+···+qn-1)(1-q)=1+q+q2+···+qn-1-(q+q2+···+qn-1+qn)=1-qn∴（＊）式成立精選ppt相減(1–q)Sn=a1-a1qn=a1(1–qn)∴當(dāng)1–q≠0,即q≠1時(shí)，Sn

a1(1–qn)1-q=當(dāng)q=1時(shí)，Sn=na1錯(cuò)項(xiàng)相減法：Sn=a1+a1q

+a1q2

+···+a1qn-1qSn=a1q

+a1q2

+···+a1qn-1+a1qn

精選ppt等比數(shù)列｛an｝前n項(xiàng)和公式為當(dāng)q≠1時(shí)Sn

a1(1–qn)1-q=當(dāng)q＝1時(shí)Sn=na1=

a1-anq1-q精選ppt練習(xí)：（1）1＋2＋4＋…＋263＝

（2）1－2＋4＋…

＋(－2)n-1=

（3）等比數(shù)列{an}中，a1=8,q=,an=,則Sn=

（4）等比數(shù)列{an}中，a1=2，S3=26,則q=

264-1

1–(-2)n3

312-4或3精選ppt例1

：求通項(xiàng)為an

=2n+2n-1

的數(shù)列的前n項(xiàng)和解：設(shè)bn=2n,且對(duì)應(yīng)的前n項(xiàng)和為

Cn=2n-1,對(duì)應(yīng)的前n項(xiàng)和為′S

n″Sn則an=bn

＋Cn

，Sn=+′S

n″Sn∴′S

n=

2(1–2n)1–2=2(2n–1)=n2∴Sn=′S

n″Sn+=2n+1+n2-2

∴″Sn=1+(2n-1

)2n精選ppt例2：求和(x+)+(x2+)+(x3+)+…+(xn+)

1y

1y2

1y3

1yn（1）當(dāng)x≠0,y≠1時(shí)（2）當(dāng)x≠0時(shí)解：當(dāng)x=1時(shí)Sn=(x+x2+…+xn)+(++…+)

1y

1y2

1yn（1）Sn=

1y(1-)

1yn1-

1y=n+

yn+1-yn

yn-1當(dāng)x≠1時(shí)Sn=

x(1-xn)1-x

1y(1-)

1yn1-

1y+

x(1-xn)1-x

yn+1-yn

yn-1+=n+(2)只須注意再討論y是否等于1的取值情況精選ppt例3：求數(shù)列：1,2x,3x2,…，nxn-1,…（x≠0)的前n項(xiàng)和解：當(dāng)x=1時(shí)Sn=1+2+3+…+n=

n(n+1)2當(dāng)x≠1時(shí)Sn=1+2x+3x2+…+nxn-1

xSn=x+2x2+…+(n-1)xn-1+nxn錯(cuò)項(xiàng)相減(1–x)Sn=1+x+x2+…+xn-1-nxn=

1-xn1-x-nxn∴Sn=

1-xn(1-x)2-

nxn1-x=

(1–x)21–(1+n)xn+xn+1綜上所述：當(dāng)x=1時(shí)Sn=

n(n+1)2當(dāng)x≠1時(shí)Sn=

(1–x)21–(1+n)xn+xn+1精選ppt等差數(shù)列{an}等比數(shù)列{an}定義an+1-an=d(常數(shù)

)an+1／

an=q(不為零的常數(shù)

)通項(xiàng)an=a1+(n–1)dan-am=(n–m)dan=a1qn-1an／

am=qn-m⑴公式⑵推導(dǎo)方法①歸納猜想驗(yàn)證法②首尾相咬累加法①歸納猜想驗(yàn)證法②首尾相咬累乘法性質(zhì)若m+n=r+s,m、n、r、s∈N*則am+an=ar+as若m+n=r+s,m、n、r、s∈N*則am·an=ar·as前n項(xiàng)和Sn⑴公式⑵推導(dǎo)方法（a1+an）nSn=

2=na1+n(n–1)

2d化零為整法當(dāng)q＝1時(shí)Sn=na1當(dāng)q≠1時(shí)Sn

a1(1–qn)1-q==

a1-anq1-q①歸納猜想驗(yàn)證法②錯(cuò)項(xiàng)相減法精選ppt方法三：Sn=a1+a2+···+an=a1+a1q+a1q2+···+a1qn-1=a1+q(a1+a1q+···+a1qn-2)=a1+qSn-1=a1+q(Sn–an)∴(1–q)Sn=a1