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数列求和大题(有答案)

2021-11-05 来源:易榕旅网
1. 已知log3x

1log23,求xx2x3xn的前n项和.

2. 设Sn=1+2+3+…+n,n∈N*,求f(n)

Sn(n32)Sn1的最大值.

3.求和:Sn13x5x27x3(2n1)xn1………………………①

4.求数列

012nn5.求证:Cn3Cn5Cn(2n1)Cn(n1)2

222,42,623,,2n2n,前n项的和.

6.求sin1sin2sin3sin88sin89的值

7.求数列的前n项和:11,

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222221a4,1a27,,1an13n2,…

8.求数列{n(n+1)(2n+1)}的前n项和.

9.求数列12,24,38,,(n

10. 求数列

112,123,,1nn1,的前n项和.

11112n),的前

n项和。

11.在数列{an}中,an的和.

12.求5,55,555,…,的前n项和。

1n12n1nn1,又bn2anan1,求数列{bn}的前n项

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答案:1解:由log3x1log23log3xlog32x12

1Snxxxx=

23nx(1x)1x12n=2(11121212n)=1-

12n

.2.解:由等差数列求和公式得Sn∴f(n)Sn(n32)Sn1n(n1),Sn(n1)(n2)

nn34n642=

1n3464n=(n18n)502150

∴当n88,即n=8时,f(n)max150

3解:由题可知,{(2n1)xn1}的通项是等差数列{2n-1}的通项与等比数列{xn1}的通项

234n之积。设xSn1x3x5x7x(2n1)x……………………….②(设制错位)234n1n(2n1)x(错位相减)再利①-②得(1x)Sn12x2x2x2x2x用等比数列的求和公式得:(1x)Sn12x(2n1)xn11xn11x(2n1)x

n∴Sn(2n1)x(1x)(1x)2n

12n.4.解:由题可知,{设Sn12Sn222222n23n}的通项是等差数列{2n}的通项与等比数列{

2n2n}的通项之积

42262624…………………………………①

………………………………②(设制错位)

24421223222222n2n1①-②得(1212n1)Sn232nn1∴Sn422n22n122n2n2n1(错位相减)

012n5.证明:设SnCn3Cn5Cn(2n1)Cn………………………….. ①

把①式右边倒转过来得

Sn(2n1)Cn(2n1)Cnnn13CnCn(反序)

10mnm01n1n又由CnCn可得Sn(2n1)Cn(2n1)Cn3CnCn………..②

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1n1nn①+②得2Sn(2n2)(Cn0CnCnCn)2(n1)2(反序相加)

∴Sn(n1)2n

6.解:设Ssin21sin22sin23sin288sin289………….①

将①式右边反序得Ssin289sin288sin23sin22sin21….②(反序)又因为sinxcos(90x),sin2xcos2x1

①+②得2S(sin21cos21)(sin=89 ∴S=44.5 7.解:设Sn(11)(1a4)(1a222cos2)(sin2289cos89)27)(1an13n2)

将其每一项拆开再重新组合得

Sn(11a1a21n1当a=1时,Snn1a(3n1)n2)(1473n2)(分组)

(3n1)n2(分组求和)

1na(3n1)n=aa1a121n当a1时,Sn1(3n1)n2

a328.解:设akk(k1)(2k1)2k3kk

nn3∴Snk(k1)(2k1)=(2kk1k13kk)

2将其每一项拆开再重新组合得

n3n2nSn=2k3kk1k1333k12k(分组)

=2(12n)3(12n)(12n) n(n1)22222=n(n1)(2n1)21118n(n1)21=

n(n1)(n2)22

解:S123(n)n242n9.

1111(123n)(23n)2222n(n1)1n2211

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10解:设an1121n123n1n1n(裂项)

则Sn1nn1(裂项求和)

=(21)(311.解: ∵an∴bn2nn122121n122)(n1n)=n11

n1nn1n2

8(1n1n1)(裂项)

∴数列{bn}的前n项和

Sn8[(1)(1n11213)(1314)(1n1n1)](裂项求和)

=8(1)=5 8nn1∴ 原等式成立

n12.解:∵an=9(10-1)

∴Sn=9(10-1)+ 9(102-1) + 9(103-1) + … + 9(10n-1) 5 =9[(10+102+103+……+10n)-n]81n+1= (10-9n-10) 55 5 5 5

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