已知{an}是公差d不等于0的等差数列,{bn}是公比

不需要分奇偶数cn=an*bn=(4-3n)*(-2)^(n-1)Sn=(4-3*1)(-2)^0+(4-3*2)(-2)^1+……+[4-3(n-1)](-2)^(n-2)+(4-3n)(-2)^(

1）∵a2=b2∴1+d=1×q∵a4=b4∴1+3d=1×q^3组合成方程组后把d=q-1带入1+3d=q^3q^3-3q+2=0q^3-3q+3-1=0q^3-1-3(q-1)=0(q-1)(q^

a2=a1+d,a3=a1+2d.,a6=a1+5d,...,a10=a1+9d,若a1,a3,a6成等比数列,则a3^2=a1*a6,(a1+2d)^2=a1*(a1+5d),得到a1=4d.则(a

a1,a3,a9成等比数列a3^2=a1*a9(a1+2d)^2=a1*(a1+8d)解得a1=d(a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13d/16d=

能看到吧?

3设第一项为a,公差为da1=aa5=a+4da17=a+16d所以a5²=a1a17所以（a+4d）²=a（a+16d）a²+8ad+16d²=a²

设等差数列为a+(n-1)d于是第一项为a第五项为a+4d第17项为a+16d根据题意第1、5、17项顺次成等比数列有（a+4d)²=a(a+16d)即a²+16ad+16d

由题知：(a1+2d)(a1+14d)=(a1+8d)^2化简得到：(a1)^2+16a1*d+28d^2=(a1)^2+16a1*d+64d^236d^2=0解得：d=0因为d≠0故无解

a9²=a15²a9²-a15²=0(a9-a15)(a9+a15)=0公差d不等于0所以a9+a15=0a1+8d+a1+14d=0a1+11d=0-----

(1)由题意a2=1+d=b2=qa6=1+5d=b3=q^2,解得：d=3,q=4.(2)由(1)知等差数列的首项为1,公差为3,所以an=1+(n-1)*3=3n-2；等比数列的首相为1,公比为4

1.S5=5a1+10d=5(a1+2d)=70a1+2d=14a3=14a7^2=a2×a22(a3+4d)^2=(a3-d)(a3+19d)a3=14代入,整理,得d(d-4)=0d=0(已知d不

因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18

再问：求k1+2k2+3k3+.......+nkn=多少再答：令S=k1+2k2+...+nkn=2*[3^0+2*3^1+3*3^2+………+n*3^(n-1)]-(1+n)n/2令T=3^0+2

(a3)^2=a13*a1(a1+2d)^2=(a1+12d)*a1d-2a1=0d=2a1s1=a1s3=3a1+3d=9a1s9=9a1+36d=81a1(s3)^2=s1*s9,所以s1s3s9

a1,a5,a17为等比数列(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+

因为1/anan+1=1/an*（an+d）=1/d[1/an-1/（an+d）]=1/d[1/an-1/an+1]所以1/a1a2+1/a2a3+…+1/anan+1=1/d[1/a1-1/a2+1

是x²-a3x+a4=0,还是x²-a3+a4=0?x^2-(a1+2d)x+(a1+3d)=0将根a1,a1+d分别代入a1^2-a1^2-2da1+a1+3d=0(2d-1)a

【解】（1）方程A(k)(X^2)+2A(k+1)X+A(k+2)=0,则其Δ=4[A(k+1)^2-A(k)*A(k+2)]=4[[A(k)+d]^2-A(k)*[A(k)+2d]]=4d^2>0;

因为{An}是等差数列,所以A2+A8=A4+A6=10,A4*A6=24,所以可将A4、A6看作方程x^2-24x+10=0的两个根,因为d

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