《全品高考復(fù)習(xí)方案》第34講數(shù)列求和第1課時(shí)分組轉(zhuǎn)化法與錯(cuò)位相減法求和1.A[解析]a1+a2+…+a10=(-1+4)+(-7+10)+…+(-25+28)=3×5=15.2.C[解析]由an+1+an=3·2n,得a1+a2=6,又a2=5,所以a1=1,所以{an}的前11項(xiàng)和為a1+(a2+a3)+(a4+a5)+(a6+a7)+(a8+a9)+(a10+a11)=1+3×(22+24+26+28+210)=4093.故選C.3.B[解析]由Sn=12+24+38+…+n2n,得12Sn=1×122+2×123+3×124+…+n·12n+1,兩式相減得12Sn=12+122+123+124+…+124.B[解析]因?yàn)閍n+1=an+2,所以an+1-an=2,則{an}是首項(xiàng)為-40,公差為2的等差數(shù)列,所以an=a1+(n-1)×2=-40+2n-2=2n-42.當(dāng)n≥21時(shí),an≥0,當(dāng)1≤n≤20時(shí),an<0,所以|a1|+|a2|+…+|a40|=-(a1+…+a20)+(a21+…+a40)=-20×(-40-2)2+20×5.B[解析]由題知Sn=22+322+423+…+n+12n,所以12Sn=222+323+424+…+n+12n+1,兩式相減可得12Sn=1+122+123+124+…+12n-n+12n+1=1+1221-12n-11-6.C[解析]在數(shù)列{an}中,a1=2,a2=0,且an+2=an+2·(-1)n,則當(dāng)n為奇數(shù)時(shí),an+2=an-2,所以{an}的奇數(shù)項(xiàng)構(gòu)成首項(xiàng)為2,公差為-2的等差數(shù)列;當(dāng)n為偶數(shù)時(shí),an+2=an+2,所以{an}的偶數(shù)項(xiàng)構(gòu)成首項(xiàng)為0,公差為2的等差數(shù)列.則S32=16×2+16×152×(-2)+16×0+16×152×2=32.故選7.AD[解析]設(shè)等差數(shù)列{an}的公差為d,由題得a1+d=4,(a1+3d)+(a1+6d)=15,解得a1=3,d=1,所以an=a1+(n-1)·d=n+2,可得bn=2n+n,所以b1+b2+b3+…+b10=(2+1)+(22+2)+(23+3)+…+(210+10)=(2+22+23+…+210)+8.210[解析]由題得a2=a4-2=4-2=2,又an+2-an=2,所以數(shù)列{an}的奇數(shù)項(xiàng)構(gòu)成首項(xiàng)為1,公差為2的等差數(shù)列,數(shù)列{an}的偶數(shù)項(xiàng)構(gòu)成首項(xiàng)為2,公差為2的等差數(shù)列,所以數(shù)列{an}的前20項(xiàng)和為a1+a2+…+a20=(a1+a3+…+a19)+(a2+a4+…+a20)=10×1+10×92×2+10×2+10×92×2=9.54-2n+54×3n[解析]因?yàn)?n+1)·an=n+13n,所以Sn=23+332+433+…+n+13n,則13Sn=232+333+…+n3n+n+13n+1,兩式相減得10.解:(1)證明:由(4an+1)an+1=3an整理得4anan+1+an+1=3an,等式兩邊同時(shí)除以an+1an,可得4+1an=則1an-2=31an+1-由a1=35,可得1a1則數(shù)列1an-2是首項(xiàng)為-13(2)由(1)得1an-2的通項(xiàng)公式為則1an=2-13n,所以a(3)由(2)知1an=2-所以Sn=2-13+2-132+…+2-13n=2n-111.BCD[解析]由3(n+1)an-nan+1=0,可得an+1n+1=3·ann,又a11=1,所以ann為等比數(shù)列,其公比為3,首項(xiàng)為1,所以ann=3n-1,所以an=n·3n-1,故B,C正確;nan=n2·3n-1,因?yàn)?n+1)an+1nan=(n+1)2·3nn2·3n-1=3(n+1)2n2不是常數(shù),所以數(shù)列{nan}不是等比數(shù)列,故A錯(cuò)誤;因?yàn)閍n=n·3n-1,所以Sn=1+2×3+3×32+…+n×3n-1①,所以3Sn=3+2×32+3×33+…+n×3n②,①-②得-2Sn=1+312.D[解析]由f(x)+f(2-x)=2,an=fn2024,得當(dāng)1≤n≤2023時(shí),an+a4048-n=fn2024+f4048-n2024=2,即a1+a4047=a2+a4046=…=a2023+a2025=2,又a2024+a2024=f20242024+f20242024=2=a1+a4047,所以2S4047=(a1+a4047)+(a2+a4046)+…+(a4047+a1)=4047(a1+a4047)=8094,所以S13.3-2n+32n[解析]∵an=2n-1,bn=2-n,∴cn=an·bn=(2n-1)·2-n,∴Sn=1×2-1+3×2-2+5×2-3+…+(2n-1)×2-n,則12Sn=1×2-2+3×2-3+5×2-4+…+(2n-1)×2-n-1,兩式相減得12Sn=2-1+2(2-2+2-3+2-4+…+2-n)-(2n-1)×2-n-1=12+2×14×1-12n-11-12-14.解:(1)因?yàn)閍1+a22+a322+…+所以當(dāng)n≥2時(shí),a1+a22+a322+…+兩式相減可得an2n-1=2,所以an=又a1=2也滿足上式,所以an=2n.(2)由(1)得bn=n·2n,所以Tn