设数列是一个公差为d的等比数列,满足前10项的和sn=110

设{A(n)}的通项公式为：A(n)=2+d(n-1){B(n)}的通项公式为：B(n)=2×q^(n-1)则{A(n)}的前n项和为：S(n)=[A(1)+A(n)]n/2=[4+d(n-1)]n/

a2=a1+d a4=a1+3d（a2）2=a1×a4即（a1+d）2=a1（a1+3d）整理得a1d=d2∵d≠0∴a1=dS10=10a1+12×10×9×d=10a1+45d=55a1

(1)令通项公式:an=a1+(n-1)da2=a1+da4=a1+3dS10=5(2a1+9d)=110由题意:a2^2=a1*a4即(a1+d)^2=a1*(a1+3d)由题意:a1=d=2所以通

(a1)(b1)=1,因b1=1,则：a1=1则：(a2)(b2)=(a1＋d)[b1q]=(1＋d)q=4,则：(1＋d)²q²=16(a3)(b3)=(a1＋2d)[b1q&#

设An=A1+(n-1)dBm=B1*q^(m-1)(此处楼上打错)因为A1=B1;A3=B3;A7=B5则可得A1(1-q^2)=2dA1(1-q^4)=6d比得q^4-3q^2+2=0(q^2-1

设a2=k,a1=k-a,a3=k+ab1*b3=b2*b2解出来k=-a/3b1=-a/3b2=2a/3b3=-4a/3公比q=-2再问：a是什么？

设an=a1*q^n-1则lgan-1+lgan+1=lga1*q^n-2+lga1*qn=lga1^2*q2n-22lgan=2lga1*qn-1=lg(a1*qn-1)^2=lga1^2*q2n-

（1）以下分段表示：a(n)=(1/4)×[2^(n-1)]=2^(n-3),1≤n≤8；a(n)=a(8)+3(n-8)=2^5+3n-24=3n+8,n＞8.（2）以下分段表示：S(n)=(1/4

(a5)^2=a1*a17(a1+4d)^2=a1(a1+16d)16d^2-8a1d=0a1=2dan通项公式为an=a1+(n-1)d=a1+(n-1)a1/2=(n+1)a1/2a5/a1=3所

A2=A1+dA4=A1+3d(A2)^2=A1×A4(A1+d)^2=A1(A1+3d)(A1)^2+2A1d+d^2=(A1)^2+3A1dA1d=d^2d≠0A1=dS10=10A1+(1/2)

证：a1,a2,a4成等比数列,则a2²=a1a4(a1+d)²=a1(a1+3d)整理,得d²-a1d=0d(d-a1)=0d≠0,要等式成立,只有d-a1=0a1=d

由等差数列有：S10=10a1+45d=165由等比数列有：(a1+d)(a1+d)=a1(a1+3d)即是a1^2+2d*a1+d^2=a1^2+a1*3dd^2=a1*d所以a1=d；代入第一式:

A2*A2=A1*A4A2=A1+dA4=A1+d得A1=dA10=10dS10=10(A1+A10)/2=110A1=d=2An=2n

（I）由a1，a2，a4成等比数列可得：(a1+2)2＝a1(6+a1)∴4=2a1即a1=2∴an=2+2（n-1）=2n（II）∵bn=n•2an，=n•22n=n•4n∴Sn＝1•4+2•42+

设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9

a2^2=a1a4(a1+d)^2=a1(a1+3d)a1^2+2a1d+d^2=a1^2+3a1da1d=d^2a1=da1=da2=d+d=2da3=d+2d=3d.an=a1+(n-1)d=d+

数列{an}是公差不为0的等差数列，设公差为d，S1，S2，S4成等比数列，则S22=S1•S4，∴（ 2a1+d）2=a1•（4a1+6d），化简可得d=2a1∴a3a1=a1+2da1=

那太简单啦,通过递推不就得了,你第一问求得的是d=2,q=3吧.由原式再往上递推一项,就有c1/b1+c2/b2+……+cn/bn+c(n+1)/b(n+1)=a(n+2),然后跟原式联立,两式相减,

∵a1=f(d－1)=(d－2)2,a3=f(d+1)=d2,∴a3－a1=d2－(d－2)2=2d,∴d=2,∴an=a1+(n－1)d=2(n－1)；又b1=f(q+1)=q2,b3=f(q－1)

设公比为q,因为a1=1,即：a(n)=q^(n-1)则：S(n)=(1-q^n)/(1-q)若{Sn}为等差数列,设公差为d则：S(n)=S(n-1)+d即：d=S(n)-S(n-1)=(1-q^n