日韩亚洲一区中文字幕,日韩欧美三级中文字幕在线,国产伦精品一区二区三区,免费在线欧美性爱链接

      1. <sub id="o5kww"></sub>
        
<legend id="o5kww"></legend>
        <style id="o5kww"><abbr id="o5kww"></abbr></style>
        <strong id="o5kww"><u id="o5kww"></u></strong>
        

        2. 


          
	
          
		
	
          
	
          
		
          
			
		
          
		
          
	
          

          



          

          

          


          

          
	
          
				
          

			
				(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(duì)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(xiàng)為1的等比數(shù)列,b1+b2+…+bn=(n∈N*).求證:{bn}是等比數(shù)列.
          

			
          
				查看答案和解析>>
			
          
		
          
				
          
									
          

          (1)Sn為等差數(shù)列{an}的前n項(xiàng)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(xiàng)a1和公比q．
          

          

          

			
          
				查看答案和解析>>
			
          
		
          
				
          
									
          

          (1)Sn為等差數(shù)列{an}的前n項(xiàng)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(xiàng)a1和公比q．
          

          

          

			
          
				查看答案和解析>>
			
          
		
          
				
          
									
          

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

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

          

          

			
          
				查看答案和解析>>
			
          
		
          
				
          
									
          設(shè){an}是正數(shù)組成的數(shù)列，其前n項(xiàng)和為Sn，且對(duì)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(xiàng)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(xiàng)和T20．
          

			
          
				查看答案和解析>>
			
          
		
          
						
          
		
          難點(diǎn)磁場(chǎng)
          6ec8aac122bd4f6e
          殲滅難點(diǎn)訓(xùn)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+1－6ec8aac122bd4f6ean}是公比為6ec8aac122bd4f6e的等比數(shù)列，且a1=6ec8aac122bd4f6e,a2=6ec8aac122bd4f6e,
          ∴an+1－6ec8aac122bd4f6ean=(a2－6ec8aac122bd4f6ea1)(6ec8aac122bd4f6e)n-1=(6ec8aac122bd4f6e－6ec8aac122bd4f6e×6ec8aac122bd4f6e)(6ec8aac122bd4f6e)n-1=6ec8aac122bd4f6e,
          ∴an+1=6ec8aac122bd4f6ean+6ec8aac122bd4f6e                                               ①
          又由數(shù)列{lg(an+1－6ec8aac122bd4f6ean)}是公差為－1的等差數(shù)列，且首項(xiàng)lg(a2－6ec8aac122bd4f6ea1)
          =lg(6ec8aac122bd4f6e－6ec8aac122bd4f6e×6ec8aac122bd4f6e)=－2,
          ∴其通項(xiàng)lg(an+1－6ec8aac122bd4f6ean)=－2+(n－1)(－1)=－(n+1),
          ∴an+1－6ec8aac122bd4f6ean=10－(n+1),即an+1=6ec8aac122bd4f6ean+10－(n+1)                                                                                                 ②
          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(xiàng)式，故可設(shè)f(x)=A(x－2a)(x－4a)(x－C),這里A、C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)(4a－C)=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
          當(dāng)p＜1時(shí),q＜1, 6ec8aac122bd4f6e
          6ec8aac122bd4f6e
          8.解：(1)an=(n－1)d,bn=26ec8aac122bd4f6e=2(n－1)d?
          Sn=b1+b2+b3+…+bn=20+2d+22d+…+2(n－1)d?
          由d≠0,2d≠1,∴Sn=6ec8aac122bd4f6e
          ∴Tn=6ec8aac122bd4f6e
          (2)當(dāng)d＞0時(shí)，2d＞1
          6ec8aac122bd4f6e
           
           
           
          		
          
	
          
	
          
		
          


          同步練習(xí)冊(cè)答案