2022英语周报九年级53答案

18.【试题情境】本题是基础性题目,属于课程学习情境,具体是数学运算学习情境【必备知识】本题考查的知识是“掌握等差数列、等比数列的通项公式与前n项和公式”【关键能力】本题考查逻辑思维能力、运算求解能力【解题思路】(1)a1·a3=a,4n=256设等比数列a,}的公比为q(q>0)2S=b62+b,-2=1时+b2-b1-2=0→b,=2n≥2时b2bn+bn)·(b,+b.>0bn1-1)=02(S.-3n)a错位相减法(2)由(1)→S解:(1)设等比数列|an}的公比为q,q>0,(1分)a12·a3=a1,a8=256,a1q·a1q=a1q,a1q=252(3分)对于2S。=b2+b。-2①当n=1时,21=2b1=b2+b1-2,解得b1=2或b1=-1(舍去)(4分)当n≥2时,2Sn=b21+bn1-2①-②得2b,=b2-b21+b。-b,1,即(b+b,1)·(b。-b1-1)=0b,+bn1>0,b一b。1-1=0,即b-b1=1,(5分)数列|b,是以2为首项,1为公差的等差数列.b,=2+(n-1)×1=n+1.(6分)(2)由(1)可得Sn=n(n+1+2)3n3)·2C(n+1)-1M,=1×22+2×2+3×24+…+(n-1)×2+nx2③则2M,=1×2+2×24+3×23+…+(n-1)×2+n×22④,(10分③-④得-M,=22+23+2++22(1-2”)n×22=-[4+(n-1)×2"2],(11分)∴M,=4+(n-1)×22(12分)解后反思》在解决数列求和问题时,首先需要判断数列通项的特征,然后选择相应的求和方法,如本题中数列{cn}的通项公式是一个等差数列和一个等比数列的通项的积,故采用错位相减法求数列cnl的前n项和

书面表达One possible versionDear PeterI'm writing to invite you to see the Chinese Painting Exhibition to be held in our city.The exhibition will start at 8: 30 am on Saturday and last till 5: 00 pm on Sunday in the city museum It'ssaid that a large number of Chinese paintings, some of which are original works by famous painters like ZhangDaqian and Qi Baishi, will be on display. Besides, there will be various souvenirs to be sold. I know you re veryinterested in traditional Chinese culture, so I can t wait to tell you the news and hope we can go together.Looking forward to your reply.You

以上就是2022英语周报九年级53答案，更多英语周报答案请关注本网站。