The series an is a series of equal differences, and the sum of the first n terms of the series bn of

1. An is a series of equal differences.

Tolerance d=(a5-a3) 2=2

The first term a1 = a3 - 2d = 1

Then the general formula for an is: an=a1+(n-1)d=2n-1

sn+bn=2, i.e., sn=2-bn

1° When n=1, then b1+b1=2, i.e., b1=1

2° When n 2, then bn=sn-s(n-1)=2-bn-[2-b(n-1)]=b(n-1)-bn

2bn=b(n-1), i.e., common ratio q=bn b(n-1)=1 2

Then the general formula for bn is: bn=b1q (n-1)=(1 2) (n-1).

2、cn=an/bn=(2n-1)*2^(n-1)

tn=1*2^0+3*2^1+5*2^2+……2n-1)*2^(n-1) ①

2tn= 1*2^1+3*2^2+……2n-3)*2^(n-1)+(2n-1)*2^n ②

Obtain: tn=1+2*2 1+2*2 2+......2*2^(n-1)-(2n-1)*2^n

1+2^2+2^3+……2^n-(2n-1)*2^n

1+2^2*[1-2^(n-2)]/(1-2)-(2n-1)*2^n

1+2^n-4-(2n-1)*2^n

2-2n)*2^n-3

1-n)*2^(n+1)-3

That is, tn=3+(n-1)*2 (n+1).

1.Let the tolerance of the equal difference series be d, which can be seen from the title:

a3=a1+2d=5a5=a1+4d=9

Solution: a1=1, d=2

an=a1+(n-1)d=1+2(n-1)=2n-1sn+bn=2,s(n-1)+b(n-1)=2 subtracted: sn+bn-s(n-1)-b(n-1)=02bn=b(n-1), bn(bn-1)=1 2 can be seen from the above equation to be proportional series, and the common ratio is 1 2;

By sn+bn=2,s1+b1=b1+b1=2,b1=1, so bn=1*(1 2) (n-1)=2 (1-n).

a1+a2+a3=4

a1+a1+d+a1+2d=4

3a1+3d=4

a3+a4+a5=10

a1+2d+a1+3d+a1+4d=10

3a1+9d=10

Stand with 3a1+3d=4.

The solution gives d = 1 a1 = 1 3

So sn=na1+d*n(n-1) 2

n/3+n(n-1)/2

3n 2-n) 6,1, it is known that the two equations in the problem are subtracted to a tolerance of 6 times. That is, 6d=6, so d=1, and because a2=4 3, a1=1 3, now we know a1, and the tolerance d, so sn can be directly represented by the formula, 0, the known number series is the first stupid number series of equal differences, and its first n terms are summed, a3+a4+a5=10

Find snrtttttttttt

When n = 1, b(1) = s(1) = (2 3) [b(1)-1] contains ruler b(1) = -2;Fraud.

When n 2, b(n) = s(n)-s(n-1)(2 3)[b(n)-1]-(2 3)[b(n-1)-1](2 3)[b(n)-b(n-1)].

then b(n) = (2) b(n-1).

So, b(n)=(2) n, this equation holds for n 1.

So. a(2)=b(1)=-2

a(5)=b(2)=4

Therefore, 3d = a(5) - a(2) = 6

That is, the tolerance of the height of the talk is d=2

Then the first term is a(1)=a(2)-d=-4

So. a(n)=-4+2(n-1)=2n-6。

According to the title, s(n) = (2 3)[b(n)-1](2 3)[(2) n-1].

1.From s1=b1=2 3(b1-1), b1=-2, and then from s2=b1+b2=2 3(b2-1), b2=-2-3b1=4, substituting b1=a2=a1+d,b2=a5=a1+4d into the two-formula air-land phase subtracts a1=-4,d=2, so an=-4+2(n-1).

2.From sn=2 3(bn-1), sn-sn-1=bn=2 3(bn-bn-1) after sorting out bn=-2bn-1, you can code the banquet to see that bn is a proportional series with a common ratio of -2, so bn=b1x(-2) (n-1)=(2)x(-2) (n-1)=(2) bucket mold is n sn=b1(1-q^n)/(1-q)=(2)(1-(-2)^n)/3

The number column is an equal difference series, and the tolerance is d

a3=5,a5=9

a1+2d=5,a1+4d=9

The solution is d=2 and a1=1

an=2n-1

The sum of the first n terms of the series is sn, and sn+bn=2 when n=1, s1=b1

2b1=2,b1=1

s(n+1)+b(n+1)=2 ②

obtain: s(n+1)-sn+b(n+1)-bn=0 s(n+1)-sn=b(n+1).

2b(n+1)=bn

b(n+1)/bn=1/2

is an equal ratio series, and the common ratio is 1 2

bn=(1/2)^(n-1)

Let the tolerance of an be k, then, a5-a3=2k=6,k=3,an=(n-3)k+a3=3n-9+5=3n-4

sn+bn=2

then, b1+b1=2, b1=1;

2b2+b1=2,b2=1/2;

sn-s（n-1）=bn；

then 2sn-s(n-1)=2;

sn=(1/2)s(n-1)+1;

s(n-1)=(1/2)s(n-2)+1;

then sn-s(n-1)=(1 2)[s(n-1)-s(n-2)];

bn=(1/2)b(n-1);

then bn is a proportional series with a common ratio of 1 2;

b1=1;bn=(1/2)^(n-1)

1.If the number column is an equal difference series, then the tolerance d=(a5-a3) (5-3)=2, and a3=a1+2d=5, we can get a1=1

So an=1+2(n-1)=2n-1

2.From sn+bn=2, s(n+1)+b(n+1)=2, and the subtraction of the two equations yields: 2b(+1)n-bn=0, that is, b(n+1)=1 2bn

When n=1, b1=1 can be obtained from sn+bn=2, and when n=2, b2=1 2 so bn=(1 2) (n-1).

an=1+2（n-1）

Bn is mathematically inducted as: bn=1 2(n-1).

2 (n-1) represents 2 to the (n-1) power.

(1) If it is known to be a series of equal differences, then: a5-a4=a4-a3=d, there is: 2a4=a3+a5=14, so, a4=7.

The tolerance is d=2, so a2=3 and a1=1.

The formula for the general term of the equal difference series is: an=a1+(n-1)d, so: an=1+2(n-1)=2n-1.

Answer: s1=b1 and s1=(1-b1) 2, solve b1=1 3=1 3 1;

s2=b1+b2 and s2=(1-b2) 2, b2=1 9=1 3 2;

s3=b1+b2+b3 and s3=(1-b3) 2, b3=1 27=1 3 3;

So, bn=1 3 n.

2) Answer: c1 = a1 · b1 = 1 3

c2=a2·b2=3/9

c3=a3·b3=5/27

cn=an·bn=(2n-1)/3^n

Therefore, tn=1 3+3 9+5 27+...2n-1) 3 n———1).

1) Divide the left and right sides of the formula by 3 to get:

1/3tn=1/9+3/27+..2n-1) 3 (n+1)-2).

1) formula - (2) formula, get:

2/3tn=1/3+2/9+2/27+..2/3^n-(2n-1)/3^(n+1)

2/3tn=2/3+2/9+2/27+..2/3^n-(2n-1)/3^(n+1)-1/3

2/3tn=(1-1/3)+(1/3-1/9)+(1/9-1/27)+.2n-1)/3^(n+1)-1/3

2/3tn=1-1/3^n-(2n-1)/3^(n+1)-1/3

tn=(3^n-n-1)/3^n

an=2n-1

bn=sn-sn-1=(bn-1-bn)/23bn=bn-1

bn=bn-1/3=b1/3^n-1

b1=s1=(1-b1)/2

b1=1/3

So, bn=1 3 n.

cn=an*bn=(2n-1)/3^n

tn=2pn-qn

where qn= 1 3 n=(1-1 3 n) 2pn= n3 n

pn/3=∑n/3^n+1

So 2pn 3=( 1 3 n)-n 3 n +12pn = (3-1 3 n-1) 2-n 3 n so tn=2pn-qn=(3-1 3 n-1) 2-n 3 n-(1-1 3 n) 2=1-(n+1) 3 n

ps: This question should be the difficulty of the last question in mathematics, and it seems that there is this question in the college entrance examination practice questions every year, which is very typical.

1.Let the tolerance be d and the tolerance be q

s15=15a1+105d=15(a1+7d)=15a8=225a8=15

a8-a3=5d=15-5=10

d=2a1=a3-2d=5-4=1

an=a1+(n-1)d=1+2(n-1)=2n-1b3=a2+a3=2×2-1+2×3-1=8b2b5=(b3/q)(b3q²)=b3²q=64q=128q=2b1=b3/q²=8/4=2

The general formula for the bn=b1q (n-1)=2 2 (n-1)=2 series is an=2n-1; The general formula for the series is bn=2.

tn=c1+c2+..cn=1×2²+2×2³+3×2⁴+.n×2^(n+1)-(2+2²+.2ⁿ)

Let cn=1 2 +2 2 +3 2 +n 2 (n + 1) then 2cn = 1 2 +2 2 +n-1)×2^(n+1)+n×2^(n+2)

cn-2cn=-cn=2²+2³+.2^(n+1)-n×2^(n+2)

4×(2ⁿ-1)/(2-1)-n×2^(n+2)=(1-n)×2^(n+2) -4

cn=(n-1)×2^(n+2) +4

tn=cn -(2+2²+.2ⁿ)

n-1)×2^(n+2) +4-2×(2ⁿ-1)/(2-1)=(2n-1)×2^(n+1) +6