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已知等差数列{An}中,A1=1,公差D>0,且A2,A5,A...

请采,可问

(1)∵a2=1+d,a5=1+4d,a14=1+13d,且a2,a5,a14成等比数列,∴(1+4d)2=(1+d)(1+13d),解得d=2,∴an=1+(n-1)?2=2n-1,又b1=a2=3,b2=a5=9,∴q=3,bn=3?3n?1=3n;(2)c1b1+c2b2+…+cnbn=an+1,即C13+C232+…+Cn3n=2n+1①,则n≥2时,C13...

数学归纳法 理解起来的话就是先证明一个易证的(这里就是证明n=1),然后证明若n=k时成立,则n=k+1时成立。连起来就是证明了n=1时成立,则n=2时成立,则n=3时成立。。。。。

(1)由题意得(1+4d)2=(1+d)(1+13d),d>0解得d=2…(3分)∴an=2n-1…(4分)又b2=a2=3,b3=a5=9,所以{bn}的公比为3,bn=3n-1…(6分)(2)∵cn=2an-18=4n-20…(7分)令cn≤0得n≤5…(9分)所以当n=4或n=5时,sn取最小值-40…(12分)

(1)∵等差数列{an}中,a1=1,公差d>0,且a2,a3+1,a4+4成等比数列,∴(2+2d)2=(1+d)(5+3d),解得d=-1(舍)或d=1,∴an=n,又b1=2,b2=4,∴bn=2n.(2)n=1时,c1a1=b1,解得c1=2,n≥2时,(c1a1+c2a2+…+cnan)-(c1a1+c2a2+…+cn?1an?1...

(1)由{bn}是等比数列,得b22=b1?b3,即(a1+4d)2=(a1+d)(a1+13d),整理得:2a1d=d2.∵a1=1,公差d>0,∴d=2.∴an=1+2(n-1)=2n-1.b1=a2=3,b2=a5=9,∴等比数列{bn}的公比q=3.∴bn=3n;(2)由c1b1+c2b2+…+cnbn=Sn,得c1b1+c2b2+…+cn?1bn?...

1、a1+a5=a2+a4=14,a2×a4=45,故a2和a4是方程x²-14x+45=0的两根,解得a2=5,a4=9,d=(9-5)÷(4-2)=2 故an=2n+1,Sn=n²+2n 2、bn=1/4n(n+1)=cn+1-cn,故cn=-1/4n

解:a1+a3+a5+.....+a17+a19=40 ......(1) a19+a17+a15+....+a3+a1 =40 ........(2) (1)+(2),得10(a1+a19)=80===>a1+a19=8===>a1+a1+18=8===>a1=-5 S19=na1+n(n-1)d/2=19*(-5)+19*18/2=19*(9-5)=76.

解: ∵a1+a5=2a3 ∴2a3+a3=-12 3a3=-12 a3=-4 a1a3a5=80 a1a5=-20 (a3-2d)(a3+2d)=-20 (-4-2d)(-4+2d)=-20 16-4d^2=-20 4d^2=36 d=3 ∴a1=a3-2d=-4-6=-10 an=-10+3(n-1)=3n-13 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 祝你学习进步,更上一层楼!...

a(n) = a + (n-1)d, d不为0. 4 = a(5) = a + 4d. 12 = 3a + 12d. 12 = a(2) + a(3) + a(k) = a + d + a+ 2d + a + (k-1)d = 3a + (k+2)d = 3a + 12d, (k+2)d = 12d, 0 = d[k+2-12] = d[k-10], k=10.

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