It is known that the number series bn antego and Sn 3 2n 2 1 2n number series an satisfy 100

b1=s1=3/2-1/2=1

n>=2: Occasionally, there is bn=sn-s(n-1)=3 2n 2-n 2-3 2(n-1) 2+(n-1) 2=3 2(2n-1)-1 2=3n-2

b1 = 3 * 1 - 2 = 1, also yes, so there is bn = 3n-2

Then there is an 3=4 - (bn+2)=4 - (3n).

i.e. an=4 (-n)=(1 4) n

cn=anbn=(1/4)^n*(3n-2)

tn=c1+c2+..cn=1/4*(1)+(1/4)^2*4+(1/4)^3*7...1/4)^n*(3n-2)

1/4tn=(1/4)^2*1+(1/4)^3*4+(1/4)^4*7+..1/4)^(n+1)*(3n-2)

tn-1/4tn=1/4+(1/4)^2*3+(1/4)^3*3+..1/4)^n*3-(1/4)^(n+1)*(3n-2)

3/4tn=1/4+3[1/16*(1-(1/4)^(n-1))/(1-1/4)]-1/4)^(n+1)*(3n-2)

1/4+1/4*(1-(1/4)^(n-1))-1/4)*(1/4)^n*(3n-2)

1/2-(1/4)^n-3n/4*(1/4)^n+1/2*(1/4)^n

1/2+1/4*(1/4)^n-3n/4*(1/4)^n

Therefore, tn=2 3+1 3*(1 4) n-n*(1 4) n

1) b1=s1=1

bn=sn-s(n-1)=3n-2,n≥2

So, bn=3n-2, n 1

The logarithm of the equation an 3=4 -(bn+2) is obtained

3log(an)=-(bn+2)log4

3log(an)=-3nlog4

log(an)=log4^(-n)

So, an=4 (-n)=(1 4) n

2) cn=an*bn=(1/4)^n*(3n-2)

tn=c1+c2+..cn

1/4+4(1/4)^2+7(1/4)^3+..3n-2)(1/4)^n

1/4)tn =(1/4)^2+4(1/4)^3+..3n-5)(1/4)^n+(3n-2)(1/4)^(n+1)

Subtract the two formulas to obtain:

tn-(1/4)tn=1/4+3(1/4)^2+3(1/4)^3+..3(1/4)^n-(3n-2)(1/4)^(n+1)

3/4)tn=1/4-(3n-2)(1/4)^(n+1)+3[(1/4)^2+(1/4)^3+..1/4)^n]

3/4)tn=1/4-(3n-2)(1/4)^(n+1)+1/4-(1/4)^n

3tn=1-(3n-2)(1/4)^n+1-(1/4)^(n-1)

tn=2/3-(n-2)(1/4)^n

bn=sn-sn-1=3n-2

Your knowledge didn't write it, right, an 3=4 -(bn+2)?? Less things.

When n 2, there is an=sn-sn-1=n2+5n-(n-1)2-5(n-1)brigade suspicion=2n 4

And after the dismantling of the hand count Wang a1 s1

an=2n＋4

sn+2n=2an

s(n-1)+2(n-1)=2a(n-1).

sn-s(n-1)+2n-2(n-1)=2an-2a(n-1)an+2=2an-2a(n-1)

an=2a(n-1)+2

Let an+c=2(a(n-1)+c).

c=2an+2=2(a(n-1)+2)

When n=1, s1=a1+2=2a1 a1=2a1+2=4

A sequence is a proportional series.

an+2=(a1+2)q^(n-1)=4*2^(n-1)=2^(n+1)

an=2^(n+1)-2

The second sub-question, ** is a part.

1/4+1/2*(1-1/2^n)-(2n+2)/2^(n+1)=3/4-(2n+3)/(2^(n+1))=3/2-(2n+3)/2^n<3/2

1. Calculating an is the key.

sn=2an-2n sn-1=2an-2(n-1) to calculate an, it is very simple to calculate according to the requirements in the next step, and the ratio should be defined according to the definition.

2. TN is expressed by bn, and it can be clearly seen that the proof result is a type of question type, and this type will be done, and the process is useless.

(1) Certificate: sn=2an-2n. 1）

s（n-1)=2a(n-1)-2(n-1) .2)

1)-(2)=2an-2n-2a(n-1)+2n-2=2an-2a(n-1)-2=sn-s(n-1)

and sn-s(n-1)=an

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

an=2a(n-1)+2

From the above equation, we can get a(n-1)=(an-2) 2

an+2)/(a(n-1)+2)=(an+2)/[(an-2)/2+2]

Simplifying the right side of the above equation yields an+2) (a(n-1)+2)=2 (constant).

It is proved that sn+2n=an

Launched s1+2*1=2a1 s1=a1

So a1=2

is a proportional sequence by the formula an=a1*q (n-1).

So an+2=(a1+2)*2 (n-1).

So an=2 (n+1)-2

2) Substituting the an in (1) into the bn.

bn=n+1

The number series is set to the number series.

Substituting an, bn gets the formula cn=(n+1) (2 (n+1)) n>=1).

Upstairs is too dedicated to not mix with drifting by.

When n=1, s1=a1=1+3a1 2

a1/2=-1

a1=-2n 2.

sn=n+3AN2 s(n-1)=(n-1)+3a(n-1).

sn-s(n-1)=an=n+3an/2 -(n-1)-3a(n-1)/2

an =3a(n-1) -2

an -1=3a(n-1)-3=3[a(n-1)-1]

an -1) [a(n-1)-1]=3, is a fixed value.

a1-1=-2-1=-3

The number series is a proportional series with -3 as the first sell and 3 as the common ratio.

an -1=(-3)×3^(n-1)=-3ⁿ

an=1-3ⁿ

The general formula for the series is an=1-3

Let the tolerance be db2=a2=1-9=-8 b20=a4=1-81=-80

b20-b2=18d=-80-(-8)=-72

d=-4b1=b2-d=-8-(-4)=-4

bn=b1+(n-1)d=-4+(-4)(n-1)=-4n

bn (an -1) = -4n old (-3 ) = 4n 3

tn=b1+b2+..bn

4(1/3+2/3²+3/3³+.n/3ⁿ)

Let cn=1 3+2 3 +3 3 +n/3ⁿ

Then cn 3 = 1 3 +2 3 +n-1)/3ⁿ+n/3^(n+1)

cn-cn/3=(2/3)cn=1/3+1/3²+.1/3ⁿ -n/3^(n+1)

1/3)(1-1/3ⁿ)/1-1/3) -n/3^(n+1)

1/2)(1-1/3ⁿ) n/3^(n+1)

cn=(3/2)(1/2)(1-1/3ⁿ) 3/2)[n/3^(n+1)]

3/4)(1-1/3ⁿ) n/(2×3ⁿ)

tn=4cn=3(1-1/3ⁿ) 2n/3ⁿ=3 -(2n+3)/3ⁿ

sn=2n^2+n

s(n-1)=2(n-1)^2+n-1

2n^2-4n+2+n-1

2n^2-3n+1

sn-s(n-1)

an=2n^2+n-2n^2+3n-1

4n-1an=4log2 (bn)+3=4n-1

4log2(bn)=4n-4

log2(bn)=n-1

bn=2^(n-1)

an*bn=(4n-1)2^(n-1)=4n*2^(n-1)-2^(n-1)

n2^(n-1+2)-2^(n-1)

n2^(n+1)-2^(n-1)

tn=1*2^2-2^0+2*2^3-2^1+..n2^(n+1)-2^(n-1)

1*2^2+2*2^3+..n2^(n+1)-(2^0+2^1+..2^(n-1))

2^0+2^1+..2 (n-1) is the first proportional sequence of k1=1 and q=2.

2^0+2^1+..2^(n-1)

1*(2^n-1)/(2-1)

2 n-1 set.

mn=1*2^2+2*2^3+..n2^(n+1)

2mn=2*2^2+2*2^4+..n2^(n+2)

Subtract the two formulas to obtain:

2mn-mn=mn

1*2^3+2*2^4+..n2^(n+2)-1*2^2-2*2^3-..n2^(n+1)

2^2-2^3-..2^(n+1)+n2^(n+2)

Number series -2 2-2 3 -..2 (n+1) is a proportional sequence with the first term = -4 and the common ratio q = 2.

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

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

4(2^n-1)

So. mn=n2^(n+2)-4(2^n-1)

n2^(n+2)-2^(n+2)+4

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

Resources. If the answer is helpful to you, I sincerely hope that you will praise it!!

Wish: Learning progress!!

sn=2n^2+2n

When n=1.

an=a1=s1=2+2=4

When n 2 and n n *.

an=sn-s(n-1)

2n 2+2n-2(n-1) 2-2(n-1)=2n 2+2n-2(n 2-2n+1)-2n+2=2n 2+2n-2n 2+4n-2-2n+2=4n is combined to obtain: an=4n(n n*).

tn=2-bn

t(n-1)=2-b(n-1)

The upper formula - the lower formula gets:

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

bn=b(n-1)-bn

2bn=b(n-1)

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

bn is a proportional series.

t1=b1=2-b1

b1=1bn=b1*q^(n-1)=1*(1/2)^(n-1)=(1/2)^(n-1)(n∈n*)

an=sn-s(n-1)

sn=3/2n^2+7/2n

s(n-1)=3/2(n-1)^2+7/2(n-1)sn-s(n-1)=3/2n^2+7/2n-[3/2(n-1)^2+7/2(n-1)]

3n-3/2+7/2

3n+2bn=2^(3n+2)

b(n-1)=2 (3n+2-3)=2 (3n+2)*1 8q=1 Yuheng8

b1=2^5

bn=32*(1/8)^(n-1)

The first n terms are transported by false letter and side wheel ta = 32 * (1-1 8 n) (1-1 8) = 28 (1-1 8 n).

Rotten Brigade 1) a1 = s1, an = sn-sn-1 that is, and finally judge whether the obtained an is satisfied with a1, if it is satisfied, it can be written as an uniformly, otherwise a1 is written separately.

2) After the first question is solved, write directly the cryptomatto out of the stove to check the bn, which is very easy to judge.