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        1. (2)Sn=a1+a2+-+an(n≥1),求Sn. 查看更多

           

          題目列表(包括答案和解析)

          已知Sn=a1+a2+…+an,n∈N*.

          (1)若Sn=n·2n-1(n∈N*),是否存在等差數(shù)列{an}對一切自然數(shù)n滿足上述等式?

          (2)若數(shù)列{an}是公比為q(q≠±1),首項為1的等比數(shù)列,b1+b2+…+bn=(n∈N*).求證:{bn}是等比數(shù)列.

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          (1)Sn為等差數(shù)列{an}的前n項和,S2=S6,a4=1,求a5

          (2)在等比數(shù)列{an}中,若a4-a2=24,a2+a3=6,求首項a1和公比q.

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          (1)Sn為等差數(shù)列{an}的前n項和,S2=S6,a4=1,求a5

          (2)在等比數(shù)列{an}中,若a4-a2=24,a2+a3=6,求首項a1和公比q.

          查看答案和解析>>

          (1)已知數(shù)列{an}的前n項和為Sn=(-1)n+1n,求通項公式an;

          (2)設(shè)數(shù)列{an}滿足1g(1+a1+a2+a3+…+an)=n+1,求an

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          設(shè){an}是正數(shù)組成的數(shù)列,其前n項和為Sn,且對于所有的正整數(shù)n,有4Sn=(an+1)2
          (I)求a1,a2的值;
          (II)求數(shù)列{an}的通項公式;
          (III)令b1=1,b2k=a2k-1+(-1)k,b2k+1=a2k+3k(k=1,2,3,…),求{bn}的前20項和T20

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          難點磁場

          6ec8aac122bd4f6e

          殲滅難點訓練

          一、1.解析:6ec8aac122bd4f6e,

          6ec8aac122bd4f6e

          答案:A

          2.解析:6ec8aac122bd4f6e

          答案:C

          二、3.解析:6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          答案:6ec8aac122bd4f6e

          4.解析:原式=6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          a?b=86ec8aac122bd4f6e

          答案:86ec8aac122bd4f6e

          三、5.解:(1)由{an+16ec8aac122bd4f6ean}是公比為6ec8aac122bd4f6e的等比數(shù)列,且a1=6ec8aac122bd4f6e,a2=6ec8aac122bd4f6e,

          an+16ec8aac122bd4f6ean=(a26ec8aac122bd4f6ea1)(6ec8aac122bd4f6e)n-1=(6ec8aac122bd4f6e6ec8aac122bd4f6e×6ec8aac122bd4f6e)(6ec8aac122bd4f6e)n-1=6ec8aac122bd4f6e,

          an+1=6ec8aac122bd4f6ean+6ec8aac122bd4f6e                                               ①

          又由數(shù)列{lg(an+16ec8aac122bd4f6ean)}是公差為-1的等差數(shù)列,且首項lg(a26ec8aac122bd4f6ea1)

          =lg(6ec8aac122bd4f6e6ec8aac122bd4f6e×6ec8aac122bd4f6e)=-2,

          ∴其通項lg(an+16ec8aac122bd4f6ean)=-2+(n-1)(-1)=-(n+1),

          an+16ec8aac122bd4f6ean=10(n+1),即an+1=6ec8aac122bd4f6ean+10(n+1)                                                                                                

          ①②聯(lián)立解得an=6ec8aac122bd4f6e[(6ec8aac122bd4f6e)n+1-(6ec8aac122bd4f6e)n+1

          (2)Sn=6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          6.解:由于6ec8aac122bd4f6e=1,可知,f(2a)=0                                                                      ①

          同理f(4a)=0                                                                                                            ②

          由①②可知f(x)必含有(x-2a)與(x-4a)的因式,由于f(x)是x的三次多項式,故可設(shè)f(x)=A(x-2a)(x-4a)(xC),這里AC均為待定的常數(shù),

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e,即4a2A-2aCA=-1                                                         ③

          同理,由于6ec8aac122bd4f6e=1,得A(4a-2a)(4aC)=1,即8a2A-2aCA=1                        ④

          由③④得C=3a,A=6ec8aac122bd4f6e,因而f(x)= 6ec8aac122bd4f6e (x-2a)(x-4a)(x-3a),

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          由數(shù)列{an}、{bn}都是由正數(shù)組成的等比數(shù)列,知p>0,q>0

          6ec8aac122bd4f6e

          p<1時,q<1, 6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          8.解:(1)an=(n-1)d,bn=26ec8aac122bd4f6e=2(n1)d?

          Sn=b1+b2+b3+…+bn=20+2d+22d+…+2(n1)d?

          d≠0,2d≠1,∴Sn=6ec8aac122bd4f6e

          Tn=6ec8aac122bd4f6e

          (2)當d>0時,2d>1

          6ec8aac122bd4f6e

           

           

           


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