《全品高考復(fù)習(xí)方案》第34講數(shù)列求和●課前基礎(chǔ)鞏固【知識(shí)聚焦】1.若干個(gè)等差或等比或可求和2.(2)并項(xiàng)法3.兩項(xiàng)之差4.積【課前演練】題組一(1)×(2)×(3)√(4)√[解析](1)錯(cuò)位相減法應(yīng)用于通項(xiàng)公式為等差數(shù)列和等比數(shù)列乘積的形式,故數(shù)列{an}不能用錯(cuò)位相減法求和.(2)當(dāng)n為偶數(shù)時(shí),Sn=2(-1+2-3+4-…+n)=2×n2=n;當(dāng)n為奇數(shù)時(shí),Sn=2(-1+2-3+4-…-n)=2-1-n-12=-(3)an=1n(n+2)=121n-(4)令S=1×12+2×122+3×123+…+9×129①,則12S=1×122+2×123+3×124+…+9×1210②,由①-②得12S=12+122+123+…+129-9×12題組二1.D[解析]依題意,an=n+1-nn(n+1)=1n-1n+1,則Sn=1-12+12-13+…+1n-1n2.45[解析]S9=a1+(a2+a3)+(a4+a5)+(a6+a7)+(a8+a9)=1+(2×2+1)+(2×4+1)+(2×6+1)+(2×8+1)=1+5+9+13+17=45.3.2[解析]由an=1n+1+n=n+1-n,得S8=a1+a2+…+a8=2-1+3-2+…+9-8=34.8194[解析]由題得S9=1×21+2×22+3×23+…+9×29,則2S9=1×22+2×23+3×24+…+9×210,兩式相減得-S9=21+22+23+…+29-9×210=2×(1-29)1-2-9×210=-8×210-2,所以S9=第1課時(shí)分組轉(zhuǎn)化法與錯(cuò)位相減法求和●課堂考點(diǎn)探究探究點(diǎn)一例1(1)C(2)26[解析](1)在數(shù)列{an}中,an=(-1)n(2n-1),則a2n-1+a2n=-(4n-3)+(4n-1)=2,所以S101=(a1+a2)+(a3+a4)+…+(a99+a100)+a101=50×2+a101=100-201=-101.故選C.(2)當(dāng)n=2k-1,k∈N*時(shí),an+2-an=2,即數(shù)列{an}的奇數(shù)項(xiàng)構(gòu)成等差數(shù)列,其首項(xiàng)為1,公差為2,則a1+a3+a5+a7+a9=5×1+5×42×2=25.當(dāng)n=2k,k∈N*時(shí),an+2an=-1,即數(shù)列{an}的偶數(shù)項(xiàng)構(gòu)成等比數(shù)列,其首項(xiàng)為1,公比為-1,則a2+a4+a6+a8+a10=1×[1-(-1)5]1-(-1)=1.所以S10=(a1+a3+a5+a7+a9)+(a對(duì)點(diǎn)演練1(1)B(2)4n+2n2n+1+4n+1-103[解析](1)由an+1+an=n+3可知,S8=(a1+a2)+(a3+a4)+(a5+a6)+(a7+a8)=4+6+8+10=(2)令an2n+1=bn,則b2b1=20563=2,所以等比數(shù)列{bn}的公比為2,所以bn=b1·2n-1=2n,所以an=2n(2n+1)=4n+2n,則Sn=a1+a2+…+an=41+21+42+22+…+4n+2n=探究點(diǎn)二例2解:(1)設(shè)等比數(shù)列{an}的公比為q(q>0),由S2=34,S4=1516可知?jiǎng)ta1(1-q2)1-將q=12代入a1(1-q2)1所以數(shù)列{an}的通項(xiàng)公式為an=a1qn-1=12(2)證明:由(1)得bn=2n則Tn=12+322+…+2n-12n,可得12Tn兩式相減可得12Tn=12+12+122+=12+121-12n-11-1則Tn=3-(2n+3)·12n<由bn=2n-12n>0,可知數(shù)列{Tn}為遞增數(shù)列,則Tn≥綜上可得12≤Tn<3對(duì)點(diǎn)演練2解:(1)設(shè)等差數(shù)列{an}的公差為d,由題意可得a1+所以an=a1+(n-1)d=4n-3.(2)設(shè)bn=an·2n,由(1)得bn=(4n-3)2n,所以Tn=1×21+5×22+9×23+…+(4n-7)2n-1+(4n-3)2n,所以2Tn=1×22+5×23+9×24+…+(4n-7)2n+(4n-3)