2024-2025學(xué)年高中數(shù)學(xué)同步備課試題 選擇性必修二（人教A版2019）第四章 4-1 第1課時(shí) 數(shù)列的概念與通項(xiàng)公式 含解析

第四章數(shù)列4.1數(shù)列的概念第1課時(shí)數(shù)列的概念與通項(xiàng)公式課后訓(xùn)練鞏固提升1.(多選題)下列數(shù)列中,既是遞減數(shù)列又是無窮數(shù)列的是()A.-3,-5,-7,-9,…B.-1,12,-13C.-1,-2,-4,-8,…D.1,2,3,4解析:A,B,C中的數(shù)列都是無窮數(shù)列,D是有窮數(shù)列,且A,C中的數(shù)列是遞減數(shù)列,故選AC.答案:AC2.已知數(shù)列3,8,13,18,…,則72A.17 B.18 C.19 D.20解析:根據(jù)題意,數(shù)列3,8,13,18,…可寫成5-2,5×2-2答案:D3.已知數(shù)列{an}的通項(xiàng)公式an=nn+1,則anan+1an+2等于(A.nn+2 B.nn+3 C.解析:因?yàn)閍n=nn+1,所以anan+1an+2=答案:B4.已知數(shù)列{an}中,an=n2+n+1,則a3=()A.4 B.9 C.12 D.13解析:數(shù)列{an}中,an=n2+n+1,則a3=32+3+1=13.答案:D5.若數(shù)列{an}的通項(xiàng)公式為an=3n2-28n+3,則數(shù)列{an}各項(xiàng)中的最小項(xiàng)是()A.第4項(xiàng) B.第5項(xiàng) C.第6項(xiàng) D.第7項(xiàng)解析:∵an=3n-又n∈N*,∴當(dāng)n=5時(shí),an最小,∴數(shù)列{an}各項(xiàng)中的最小項(xiàng)是第5項(xiàng).答案:B6.(多選題)若數(shù)列{an}的前4項(xiàng)依次是2,0,2,0,則這個(gè)數(shù)列的通項(xiàng)公式可以是()A.an=1+(-1)n+1 B.an=1-cosnπC.an=2sin2nπD.an=1+(-1)n-1+(n-1)(n-2)解析:將前4項(xiàng)代入選項(xiàng)檢驗(yàn)可得只有D不符合.答案:ABC7.已知數(shù)列{an}的通項(xiàng)公式為an=1n(n+2),則a10=,若an=1168解析:∵an=1n(n+2),∴由an=1n(n+2)=1168,得n2+2n-168=0,得答案:11208.觀察下列數(shù)列的特點(diǎn),用一個(gè)適當(dāng)?shù)臄?shù)填空:112,245,3910,,52526解析:觀察數(shù)列各項(xiàng)知,an=n+n2n2+1,n∈N*,答案:4169.已知{an}是遞增數(shù)列,且an=n2+λn,則實(shí)數(shù)λ的取值范圍是.

解析:設(shè)f(n)=an=n2+λn,其圖象的對(duì)稱軸為直線n=-λ2,要使數(shù)列{an}為遞增數(shù)列,只需使定義在正整數(shù)上的函數(shù)f(n)為增函數(shù),故只需滿足-λ2<32,答案:(-3,+∞)10.根據(jù)數(shù)列的前幾項(xiàng),寫出下列各數(shù)列的一個(gè)通項(xiàng)公式:(1)45,1(2)12,2,92,8,252(3)1,3,6,10,15,…;(4)7,77,777,….解:(1)注意數(shù)列前4項(xiàng)中有兩項(xiàng)的分子為4,不妨把分子統(tǒng)一為4,即為45,48,411,414,(2)把分母統(tǒng)一為2,則有12,42,92,16(3)注意6=2×3,10=2×5,15=3×5,規(guī)律還不明顯,再把各項(xiàng)化為分母為2的分?jǐn)?shù),即1×22,2×32,(4