已知首项为1的等差数列an,其公差大于0,且a3,a7 2.3a
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(1)对等差数列an,有Sn=a1+n*(n-1)d/2=1+n*(n-1)d/2对等比数列bn,有Tn=b1*(1-q^n)/(1-q)=(1-q^n)/(1-q)又a8=a1+7d=1+7d,b3
∵a1+b1=5,a1,b1∈N*,∴a1,b1有1和4,2和3,3和2,4和1四种可能,当a1,b1为1和4的时,c1=ab1=4,前10项和为4+5+…+12+13=85;当a1,b1为2和3的时
∵Sn=4n+n(n−1)2×4=2n2+2n,∴1Sn=12n2+2n=12(1n−1n+1).∴数列 {1Sn}的前n项和=12[(1−12)+(12−13)+…+(1n−1n+1)]=
缺少条件,{an}为正项数列,否则log3(an)无意义,题目没法解.证:数列为正项数列,公比q>0a(n+1)/an=qb(n+1)-bn=log3[a(n+1)]-log3(an)=log3[a(
数列{Sn/n}构成一个公差为2的等差数列,∴Sn/n=2n,∴Sn=2n^2,∴a3=S3-S2=18-8=10.
a1,a7,a4成等差数列2a7=a1+a42a1q^6=a1+a1q^32q^6=1+q^32q^6-q^3-1=(2q^3+1)(q^3-1)=0因为公比Q不等于1,所以,q^3=-1/2,2S3
a1,a7,a4成等差数列2a7=a1+a42a1q^6=a1+a1q^32q^6=1+q^32q^6-q^3-1=(2q^3+1)(q^3-1)=0因为公比Q不等于1,所以,q^3=-1/2,2S3
你县假设An=1+(n-1)*1Bn=4+(n-1)*1则Cn=A(n+3)下角标n+3是由Bn整理的
a1+2d=11(a1+a1+8d)*9/2=153∴a1=5d=3∴an=5+3(n-1)=3n+2
设首项为a1,方差为da1=a3-2d=11-2d,a9=a3+6d=11+6dS9=n(a1+a9)/2=9*(11-2d+11+6d)/2=153d=3a1=a3-2d=11-2d=5通项公式=a
s9=9a1+9×8÷2×d=1539a1+36d=153a1+4d=17a1+2d=11所以a1=5d=3所以an=a1+(n-1)d=5+3(n-1)=3n+2
S9=a1+a2+a3+a4+a5+a6+a7+a8+a9=(a3-2d)+(a3-d)+a3+(a3+d)+(a3+2d)+(a3+3d)+(a3+4d)+(a3+5d)+(a3+6d)=9*a3+
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+
∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1(1+12+…+132)=a1•1−1261−12=38,∴a1
n=1+1/an=1+1/(a+n-1),1/(a+n-1)是反比例函数,渐近线X=1-a,Y=1,8小于(1-a)小于9,所以-8小于a小于-7
设an=a1+(n-1)d=10+(n-1)dSn=na1+(n-1)nd/2=10n+(n-1)nd/2S12=120+66d=-125那么d就算出来了d=-245/66所以an=10+(n-1)(
(1)证明:∵n、an、Sn成等差数列∴2an=n+Sn,∴2(Sn-Sn-1)=n+Sn,∴Sn+n+2=2[Sn-1+(n-1)+2]∴Sn+n+2Sn−1+(n−1)+2=2∴{Sn+n+2}成
An/n=1+(n-1)d/2Bn=(1-q^n)/(1-q),Sn=n+(n-1)q+……+q^(n-1)Sn*q-Sn=-n+q+q^2+……+q^n=>Sn=n/(1-q)-q(1-q^n)/(
易得An=½n(1+(n-1)d);Bn=(1﹣q^n)/(1-q);所以Sn=n/(1-q)﹣(q﹣q^(n+1))/(1-q)²所以An/n﹣Sn=½(1+(n-1)