等差数列{an}的首项为1,其前n项和a1 a2 ... an
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∵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)]=
设{an}公差为d,{bn}公比为q,q>0a3+b3=a1+2d+b1q^2=1+2d+3q^2=17d=(16-3q^2)/2T3-S3=b1(1+q+q^2)-(a1+a1+d+a1+2d)=3
an=a1+(n-1)d=2n+1Sn=[n(a1+an)]/2=n(n+2)1/Sn=1/n(n+2)=[1/n-1/(n+2)]/2An=S1+S2+...+Sn=21/40-(2n+3)/2(n
an=1/n(n+1)就不是等差数列.an=1/n(n+1)=1/n-1/(n+1),所以S10=1-1/2+1/2-1/3+...+1/10-1/11=10/11
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
S(Cn)(10)=A(B1)+A(B2)+A(B3)+A(B4)+A(B5)+A(B6)+A(B7)+A(B8)+A(B9)+A(B10)由于{Bn}是公差为1的等差数列,所以A(B1),A(B2)
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+
an=2+(n-1)=n+1bn=2^(an)=2^(n+1)Tn=b1+b2+...+bn=2(2^1+2^2+...+2^n)=4(2^n-1)/(2-1)=2^(n+2)-4
S14=(a1+a14)*7=98a1+a14=14a1+a1+13d=142a1+13d=14(1)a11=a1+10d=0(2)所以d=-2,a1=20an=20-2(n-1)=-2n+22(2)
an=2+(n-1)*1=1+nSn=(2+an)*n/2=(3+n)*n/2
∵{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
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
24=S4=a1+a2+a3+a4=2(a2+a3)=>a2+a3=12a2*a3=35=>a2=5,a3=7=>a1=3=>an=3+(n-1)*2=2n+1bn=1/an*a(n+1)=1/((2
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)
an=a1+(n-1)d=2+(n-1)*1=n+1Sn=a1n+n(n-1)*d/2=2n+n(n-1)/2=2n+n^2/2-n/2=(n^2+3n)/2