Let Xn be a proportional series where all terms are positive, Yn be a series of equal differences, a

q^4+2d=20 (2)

q^2-2q^4=-28

2q^4-q^2-28=0

2q²+7)(q²-4)=0

q>0q=2d=2

xn=1*2^(n-1)=2^(n-1)

yn=1+2(n-1)=2n-1

2) xiyj=(2i-1)*2 (j-1) The sum of all such terms is.

s=[1+3+5+..2n-1)]*1+2+4+..2^(n-1)](1+2n-1)*(n/2)*(2^n -1)n²*(2^n -1)

x3+y5=13 x5+y3=21 x1=y1=1 (let q be the common ratio value, d is the tolerance value).

That is, x1*q*q+y1+4d=13 x1*q*q*q*q+y1+2d=21 and xn are positive, and we get: q=2 d=2

Therefore, xn=2 (n-1) [2 is the base, n-1 is the exponent, that is, it is the (n-1) power of 2].

yn=2n-1

s=xi*yj=2^（i-1）*(2j-1)=j*（2^i）-2^（i-1）

314444 = 6555555466, and there are 6555555446 seconds left until the end of the question.

Because the three numbers are in a series of equal differences, we can get x+z=2y, so there is y+2y=18, and the solution is y=18 3=6.

Because x, y, and z are the difference series of holes, there is 2*y=x+z, and Jingxiao knows x+y+z=18, and from these two conditions, we can know that 3*y=18, and the bright trembling draft is y=6.

1. Proof.

Let xn=x1*b (n-1) x1 be a positive integer that is not equal to 1, and b be the common ratio yn=2 axn=2(logax1+(n-1)logab), so yn=2logax1+(n-1)(2*logab) is the first term is: 2logax1, and the tolerance is: 2*logab.

2. According to y4=17, y7=11, so the tolerance = -2 first term = 23 = > y11 = 1 y12 = -1 = > s11 is the maximum, the maximum is: (23 + 1) * 11 2 = 132

1. Prove that the shed is cheating orange.

Let xn=x1*b (n-1).

x1 is a positive integer that is not equal to 1, and b is a common ratio.

yn=2 axn=2(logax1+(n-1)logab) so yn=2logax1+(n-1)(2*logab) is the first difference series with the terms :2logax1 and the tolerance is:2*logab.

2. According to y4=17, y7=11, so the tolerance = -2 first chain group item = 23> y11 = 1

y12=-1

S11 is the largest, and the maximum is: (23+1)*11 2=132

yn logxn=2(a>0 and a≠1), 18 logx3=12 logx6=2, logx3=9,logx6=6, let the common ratio of the proportional series {xn} be q, then.

logx1+2logq=9, logx1+5logq=6, the solution yields logq=-1, logx1=11, q=1 a, x1=a 11, xn=a 11*a (1-n)=a (12-n), substitution, yn=2(12-n)=24-2n

1) Let the first m term of the sequence {yn} be the maximum, then ym>=0>y, 12-n>=0>11-n, n + xn 0 at 111, there is no m satisfying the problem set;

01,<==>12-n<0,<==>n>12, take m=12.

3）an=log(x)

log(x)/logxn

11-n)/(12-n)

1+1 (n-12)(n>13,n n), is a subtractive function, an>a

Solution: (1) Let the common ratio of the series be q, then xn=x1qn-1, yn=2logaxn=2logax1+2(n-1)logaq, yn-1-yn=2logax1+2nlogaq-[2logax1+2(n-1)logaq]=2logaq are constants, which are equal difference series;

2) Let the tolerance be d, from y4=17, y7=11, you can get y1+3d=17, y1+6d=11

The solution is y1=23, d=-2, yn=23+(n-1) (2)=25-2n, let the sum of the first n terms be tn, tn=n(23+25-2n)2=-n2+24n=-(n-12)2+144, when n=12, the sum of the first 12 terms is maximum, and the maximum value is 144

1. If it is proportional, then the number column is equal difference, and the tolerance d=[y6 y3] 3= 2, the first term y1=22, yn=24 2n, so that the sum of the first 11 terms or the first 12 terms is the maximum (y12=0, all negative terms from 13 terms), and the maximum value is 132;

2. It does not exist. xn = a (12 n), if 011 is greater than 1. In the same way, the opposite is true for A> Shenqi1, so there is no m;

3. The base of the logarithmic form of an is xn=a (12 n), and the true number is x(n 1)=a (11 n), then an=(11 n) (12 n)=[n 12) 1] (n 12)=1 1 (n 12), we can see that when the dan filial piety pants n>13, an is decreasing.

x(n)=1+(n-1)d, d is not 0

y(n)=d^(n-1),x(6)=1+5d=y(3)=d^2, 0= d^2 - 5d - 1. delta=25+4=29.d = [5 + (29) (1 2)] 2 or d = [5 - (29) (1 2)] 2

If d>0 d=[5+(29)^(1/2)]/2.

log [y(n)]=(n-1)log (d), let d = log (d), 1=(1 d) log (d)=log [d (1 d)],a=d (1 d)

b=1 then log (d) = log [a d] = d, log [y(n)] + b = (n-1) log (d) + 1 = (n - 1) d + 1 = x(n)

a,b,a=d (1 d),b=1,d=[5+(29) (1 2)] 2

Hmmm·· This problem can be solved like this (d is the tolerance and q is the common ratio) Note: qn is the nth power of q.

x1=y1 (1)

x1+d=y1*q (2)

x1+5d=y1*q2 (3)

Simultaneous (1) (2) has x1=d (q-1) brought in (3) has d (q-1)+5d=dq2 (q-1).

1+5*(q-1)=q2 after the approximation

q<>1 gives q=4

x1=d (q-1) and x1=1 so d=3xn=1+3*(n-1)=3n-2

yn=4 to the power (n-1).

You can know that a=4 (1 3) to the power b=1 - the answer is a bit weird... That's the idea.

Let's assume that there exists, find ab, and then find the general term, and substitute the logarithm of xn=log, with a as the base, and yn + b to see if it can be satisfied.

1) (xn) an=(xn+1) an+1=(xn+2)an+2=k gives xn=k (1 an),x(n+1)=k (1 a(n+1)),x(n+2)=k (1 (an+2)).

From the proportional stimulus match sequence {xn}, it can be seen that (xn+1) 2=xn*xn+2 k (2 a(n+1))=k (1 an)*k (1 (an+2)).

2 a(n+1)=1 an+1 (an+2) is the difference series.

2) From 1 a1 = 1, 1 a8 = 15 to obtain an = 1 (2n-1). Let sn=a(n+1)+a(n+2)+....a2n

s (n+1)=(n+2)+…a2n+a(2n+1)+2n(2n+2),s(n+1)-sn=a(2n+1)+a(2n+2)-a(n+1)>0 constant (the operation is saved) the smallest value of sn is s2=12 35

The original inequality can be reduced to logm+1(x)1

Give more points.