an等差公差d不等于0
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数列an满足条件:A1=1,A2=r(r>0)数列{an+an+1}是公差为d的等差数,令bn=an+an+1即首项b1=a1+a2=1+rb3=a3+a4=b1+2d=1+r+2db5=a5+a6=
额..a1=da2=2dq=a2/a1=2d/d=2.
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=
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
已知公差为d(d不等于0),a1=1,那么:a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d又a2a5a14依次成等比数列,所以:(a5)²=a2*a14
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.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
an=Sn-Sn-1=2^n-1-{2^(n-1)-1}=2x2^(n-1)-1-2^(n-1)+1=2^(n-1)a1=s1=1,所以an通项公式为2^(n-1)b1=a1=1,b3=1+2d,b9
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a1,a2,a5成等比数列,根据等比中项公式:a2^2=a1xa5(1+d)^2=1x(1+4d)d^+2d+1=4d+1d^2-2d=0d=0或d=2因为d不等于0,所以d=2
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
【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;
是A1,A3,A4等比数列吧?∵A1,A3,A4等比数列∴(a3)²=(a1)×(a4)(a1+2d)²=(a1)(a1+3d)a²₁+4d²+4a
证明:左边=1/(a1a2)+1/(a2a3)+...+1/(an-1*an)=1/d(1/a1-1/a2)+1/d(1/a2-1/a3)+...+1/d(1/an-1-1/an)=1/d[(1/a2
令an=a1+(n-1)*d由题意:a1+4d=10a1+11d=31解得:d=3a1=-2很高兴为你解决问题!