板塊二數(shù)列微專題9數(shù)列求和的常用方法高考定位近幾年高考,數(shù)列求和常出現(xiàn)在解答題的第(2)問,主要考查通過分組轉(zhuǎn)化、錯位相減、裂項相消等方法求數(shù)列的和,難度中檔.高考定位【

真題體驗

】1.(2025·天津卷)已知數(shù)列{an}的前n項和Sn=-n2+8n,則{|an|}的前12項和為 A.48 B.112 C.80 D.114√當(dāng)n=1時,a1=S1=-1+8=7,當(dāng)n≥2時,an=Sn-Sn-1=-n2+8n-[-(n-1)2+8(n-1)]=-2n+9,顯然a1=7也符合該式,所以an=-2n+9,所以當(dāng)n≤4時,an>0,當(dāng)n≥5時,an<0,所以{|an|}的前12項和為|a1|+|a2|+|a3|+…+|a12|=a1+a2+a3+a4-(a5+a6+…+a12)=2(a1+a2+a3+a4)-(a1+a2+…+a12)=2S4-S12=2(-16+32)-(-122+8×12)=80,故選C.

(2)給定正整數(shù)m,設(shè)函數(shù)f(x)=a1x+a2x2+…+amxm,求f'(-2).

精準(zhǔn)強(qiáng)化練熱點(diǎn)一分組求和與并項求和熱點(diǎn)二裂項相消法求和熱點(diǎn)三錯位相減法求和熱點(diǎn)突破已知數(shù)列{an}的前n項和為Sn,a1=2,a2=4,且Sn+2-2Sn+1+Sn=2.(1)證明:數(shù)列{an}是等差數(shù)列,并求{an}的通項公式;熱點(diǎn)一分組求和與并項求和例1由Sn+2-2Sn+1+Sn=2,得Sn+2-Sn+1-(Sn+1-Sn)=2,∴an+2-an+1=2,又a2-a1=4-2=2,∴數(shù)列{an}是以2為首項,2為公差的等差數(shù)列,∴an=2n.(2)若等比數(shù)列{bn}滿足b1=1,b2+b3=0,求數(shù)列{an·bn}的前2n項和T2n.設(shè)等比數(shù)列{bn}的公比為q,q≠0,則b2+b3=q+q2=0,∴q=-1,∴bn=(-1)n-1,∴an·bn=2n·(-1)n-1,∴T2n=2-4+6-8+…+2(2n-1)·(-1)2n-2+2(2n)·(-1)2n-1=(2-4)+(6-8)+…+[2(2n-1)·(-1)2n-2+2(2n)·(-1)2n-1]=-2+(-2)+…+(-2)=-2n.

規(guī)律方法(2025·棗莊二模)在數(shù)列{an}中,a1=20,an+1=an+2n-2.(1)求{an}的通項公式;訓(xùn)練1

(2)若bn=an-2n,求數(shù)列{bn}的前n項和Sn的最大值.

(2025·沈陽模擬)已知等差數(shù)列{an}的首項a1=1,公差為d,記{an}的前n項和為Sn,S4-2a2a3+14=0.(1)求數(shù)列{an}的通項公式;熱點(diǎn)二裂項相消法求和例2由題意可得,S4-2a2a3+14=4a1+6d-2(a1+d)(a1+2d)+14=4+6d-2(1+d)(1+2d)+14=0,整理得d2=4,則d=±2,可得an=1+2(n-1)=2n-1或an=1-2(n-1)=-2n+3,故an=2n-1或an=-2n+3.

規(guī)律方法

訓(xùn)練2

(2025·合肥二模)已知{an}是等差數(shù)列,{bn}是等比數(shù)列,且a1=b1=3,a2+a4=2b2,a1a3=b3.(1)求{an}和{bn}的通項公式;熱點(diǎn)三錯位相減法求和例3設(shè)公差為d,公比為q(q≠0),a2+a4=2b2,故2a1+4d=2b1q,6+4d=6q,a1a3=b3,故3(3+2d)=3q2,故an=3+3(n-1)=3n,bn=3·3n-1=3n.

用錯位相減法求和時,應(yīng)注意:(1)等比數(shù)列的公比為負(fù)數(shù)的情形;(2)作差后所得等比數(shù)列的項數(shù);(3)最后一項的符號.易錯提醒(2025·郴州三模)已知數(shù)列{an}為等差數(shù)列,且a2=3,a4+a5+a6=27.(1)求數(shù)列{an}的通項公式;訓(xùn)練3

(2)已知數(shù)列{bn}的前n項和為Sn,且Sn=2bn-2,求數(shù)列{bn}的通項公式;當(dāng)n=1時,b1=S1=2b1-2,解得b1=2,當(dāng)n≥2且n∈N*時,由Sn=2bn-2得Sn-1=2bn-1-2,上述兩個等式作差可得bn=2bn-2bn-1,可得bn=2bn-1,所以數(shù)列{bn}是首項和公比均為2的等比數(shù)列,故bn=2×2n-1=2n.(3)已知數(shù)列{cn}滿足:cn=an·bn,求數(shù)列{cn}的前n項和Mn.

【精準(zhǔn)強(qiáng)化練】1.(2025·昆明診斷)已知數(shù)列{an}的前n項和為Sn,5Sn+an=1. (1)求數(shù)列{an}的通項公式;

4.(2025·南京段測)已知{an}是各項均為正數(shù),公差不為0的等差數(shù)列,其前n項和為Sn,且