将数列2n-1与3n-2
@童学2570：数列an=2n - 1,bn=3n - 2,将集合aub中的元素从小到大依次排列,构成数列c1c2c3,求cn的通项公式 - 作业帮
鞠聂13965817438…… [答案] 显然an的项均为奇数,而bn=3n-2 故只需n为奇数即可.则cn=3(2n-1)-2=6n-5
@童学2570：设数列An=2n - 1,Bn=3n+2由数列{An}与{Bn}相同项构成新数列记为{Cn}.求{Cn}前30项之和. -
鞠聂13965817438…… 易得Cn=6n-1,C1=5,C30=6*30-1,前三十项之和就是2760.
@童学2570：已知:an=3n - 2,bn=a^(2n - 1),求数列{anbn}的前n项和 -
鞠聂13965817438…… .+a^(2n-1)) a^3+a^5+a^7+......+a^(2n-1)是等比数列解:设Cn=anbn=(3n-2)a^(2n-1), 则Sn=a+4a^3+7a^5+10a^7+……+(3n-5)a^(2n-3)+(3n-2)a^(2n-1), ① 两边同乘以a^2 得a^2Sn=a^3+4a^5+7a^7+……+(3n-5)a^(2n-1)+(3n-2)a^(2n+1), ② 两式错位相减(1-a^2)Sn=a+(4-1)a^3+(7-4)a^5+(10-7)a^7+....+(3n-2)a^(2n+1), =a+(3n-2)a^(2n+1)+3(a^3+a^5+a^7+
@童学2570：求数列前n项的和
鞠聂13965817438…… a(n)=(2n-1)+…+(3n-2)=n[(2n-1)+(3n-2)]/2=(5/2)n²-(3/2)n S(n)=∑a(n)=(5/2)∑n²-(3/2)∑n=(5/2)[n(n+1)(2n+1)/6]-(3/2)[n(n+1)/2]=n(n+1)(5n-2)/6
@童学2570：数列1*n,2(n - 1),3(n - 2),…,n*1的和为? -
鞠聂13965817438…… Sn=1*n+2(n-1)+3(n-2)…+(n-1)*2+n*1 =n+2n-1*2+3n-2*3+……+n^2-(n-1)n =n+2n+3n+……n^2-(1*2+2*3+……(n-1)n) =n(1+2+3+……+n)-(1/3)(1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+……+(n-1)n(n+1)-(n-2)(n-1)n) =n*n*(n+1)/2-(1/3)(n-1)n(n+1) =n(n+1)(n/2-1/3n+1/3) =n(n+1)(n+2)/6 Sn=n(n+1)(n+2)/6
@童学2570：已知数列{an}的前n项和公式为sn=2n^2 - 3n - 1,则通项公式an=? - 作业帮
鞠聂13965817438…… [答案] sn=2n^2-3n-1 s(n+1)=2(n+1)^2-3(n+1)-1 =2(n^2+1+2n)-3n-4 =2n^2+2+4n-3n-4 =2n^2+n-2 s(n-1)=2(n-1)^2-3(n-1)-1 =2(n^2+1-2n)-3n+2 =2n^2+2-4n-3n+2 =2n^2-7n+4 an=sn-s(n-1)=4n-5 a(n+1)=s(n+1)-sn=4n-1=4(n+1)-5 所以通向公式an=4n-5
@童学2570：已知数列 {an} 的前N项和为Sn=3n^2+2n - 1 求an -
鞠聂13965817438…… an={4 (n=1) 6n-1 (n≥2 且n属于N) S1=a1=4 S(n-1)=3n^2-4n an=Sn-S(n-1)=6n-1 代入1,n=1时不满足此式 所以分开写
@童学2570：求数列{2的n次方分之1+3n - 2}前n项的和. -
鞠聂13965817438…… an=1/2^n+3n-2 前n项和Sn =(1/2+1/2^2+1/2^3+……+1/2^n)+3(1+2+3+……+n)-2n =1/2*(1-1/2^n)/(1-1/2)+3*n(n+1)/2-2n Sn=1-1/2^n+n(3n-1)/2 解毕
@童学2570：{an}{bn}等差,前n项和的比为Sn/Tn=(2n+1)/(3n - 2),求lim an/bn - 作业帮
鞠聂13965817438…… [答案] Sn/Tn=(2n+1)/(3n-2),又:an/bn=S(2n-1)/T(2n-1)=[2(2n-1)+1]/[3(2n-1)-2] =[4n-1]/[6n-5] 则:lim an/bn=2/3
@童学2570：求下列数列的通项公式,Sn是其前n项和.(1)Sn=2n2 - 3n - 1;(2)Sn=3n - 2n+1 -
鞠聂13965817438…… (1)n=1时,a1=s1=2-3-1=-2,n≥2时,an=Sn-Sn-1=(2n2-3n-1)-[2(n-1)2-3(n-1)-1]=4n-5,经检验当n=1时,4n-5=-1≠-2,∴an= ?2 ,n=1 4n?5 ,n≥2 . (2)n=1时,a1=s1=3-2+1=2,n≥2时,an=Sn-Sn-1=3n-2n+1-[3n-1-2(n-1)+1]=2*3n-1-2,经检验当n=1时,2*3n-1-2=0≠2,∴an= 2 ,n=1 2*3n?1?2 ,n≥2 .
鞠聂13965817438…… [答案] 显然an的项均为奇数,而bn=3n-2 故只需n为奇数即可.则cn=3(2n-1)-2=6n-5
@童学2570：设数列An=2n - 1,Bn=3n+2由数列{An}与{Bn}相同项构成新数列记为{Cn}.求{Cn}前30项之和. -
鞠聂13965817438…… 易得Cn=6n-1,C1=5,C30=6*30-1,前三十项之和就是2760.
@童学2570：已知:an=3n - 2,bn=a^(2n - 1),求数列{anbn}的前n项和 -
鞠聂13965817438…… .+a^(2n-1)) a^3+a^5+a^7+......+a^(2n-1)是等比数列解:设Cn=anbn=(3n-2)a^(2n-1), 则Sn=a+4a^3+7a^5+10a^7+……+(3n-5)a^(2n-3)+(3n-2)a^(2n-1), ① 两边同乘以a^2 得a^2Sn=a^3+4a^5+7a^7+……+(3n-5)a^(2n-1)+(3n-2)a^(2n+1), ② 两式错位相减(1-a^2)Sn=a+(4-1)a^3+(7-4)a^5+(10-7)a^7+....+(3n-2)a^(2n+1), =a+(3n-2)a^(2n+1)+3(a^3+a^5+a^7+
@童学2570：求数列前n项的和
鞠聂13965817438…… a(n)=(2n-1)+…+(3n-2)=n[(2n-1)+(3n-2)]/2=(5/2)n²-(3/2)n S(n)=∑a(n)=(5/2)∑n²-(3/2)∑n=(5/2)[n(n+1)(2n+1)/6]-(3/2)[n(n+1)/2]=n(n+1)(5n-2)/6
@童学2570：数列1*n,2(n - 1),3(n - 2),…,n*1的和为? -
鞠聂13965817438…… Sn=1*n+2(n-1)+3(n-2)…+(n-1)*2+n*1 =n+2n-1*2+3n-2*3+……+n^2-(n-1)n =n+2n+3n+……n^2-(1*2+2*3+……(n-1)n) =n(1+2+3+……+n)-(1/3)(1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+……+(n-1)n(n+1)-(n-2)(n-1)n) =n*n*(n+1)/2-(1/3)(n-1)n(n+1) =n(n+1)(n/2-1/3n+1/3) =n(n+1)(n+2)/6 Sn=n(n+1)(n+2)/6
@童学2570：已知数列{an}的前n项和公式为sn=2n^2 - 3n - 1,则通项公式an=? - 作业帮
鞠聂13965817438…… [答案] sn=2n^2-3n-1 s(n+1)=2(n+1)^2-3(n+1)-1 =2(n^2+1+2n)-3n-4 =2n^2+2+4n-3n-4 =2n^2+n-2 s(n-1)=2(n-1)^2-3(n-1)-1 =2(n^2+1-2n)-3n+2 =2n^2+2-4n-3n+2 =2n^2-7n+4 an=sn-s(n-1)=4n-5 a(n+1)=s(n+1)-sn=4n-1=4(n+1)-5 所以通向公式an=4n-5
@童学2570：已知数列 {an} 的前N项和为Sn=3n^2+2n - 1 求an -
鞠聂13965817438…… an={4 (n=1) 6n-1 (n≥2 且n属于N) S1=a1=4 S(n-1)=3n^2-4n an=Sn-S(n-1)=6n-1 代入1,n=1时不满足此式 所以分开写
@童学2570：求数列{2的n次方分之1+3n - 2}前n项的和. -
鞠聂13965817438…… an=1/2^n+3n-2 前n项和Sn =(1/2+1/2^2+1/2^3+……+1/2^n)+3(1+2+3+……+n)-2n =1/2*(1-1/2^n)/(1-1/2)+3*n(n+1)/2-2n Sn=1-1/2^n+n(3n-1)/2 解毕
@童学2570：{an}{bn}等差,前n项和的比为Sn/Tn=(2n+1)/(3n - 2),求lim an/bn - 作业帮
鞠聂13965817438…… [答案] Sn/Tn=(2n+1)/(3n-2),又:an/bn=S(2n-1)/T(2n-1)=[2(2n-1)+1]/[3(2n-1)-2] =[4n-1]/[6n-5] 则:lim an/bn=2/3
@童学2570：求下列数列的通项公式,Sn是其前n项和.(1)Sn=2n2 - 3n - 1;(2)Sn=3n - 2n+1 -
鞠聂13965817438…… (1)n=1时,a1=s1=2-3-1=-2,n≥2时,an=Sn-Sn-1=(2n2-3n-1)-[2(n-1)2-3(n-1)-1]=4n-5,经检验当n=1时,4n-5=-1≠-2,∴an= ?2 ,n=1 4n?5 ,n≥2 . (2)n=1时,a1=s1=3-2+1=2,n≥2时,an=Sn-Sn-1=3n-2n+1-[3n-1-2(n-1)+1]=2*3n-1-2,经检验当n=1时,2*3n-1-2=0≠2,∴an= 2 ,n=1 2*3n?1?2 ,n≥2 .
