The first term of the known sequence an is a1 5,

1、sn+1=2sn+10-n①

So when n 2, replace n with n-1 to get sn=2sn-1+10-(n-1) -a(n+1)=2an-1

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

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

So it's a proportional sequence with the first term of 4 and a common ratio of 2 (you have to remember the format, otherwise you will be deducted) 2, an-1=4 2 (n-1)=2 (n+1), so an=2 (n+1)+1

Then use the grouping summation method. 2 (n+1) is proportional and 1 is a constant sequence.

So sn=a1(1-q n) (1-q)+n=5(1-2 n) (1-2)+n=5(2 n-1)+n

s(n+1)=2sn+10-n

sn=2s(n-1)+11-n

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

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

So it's a proportional series.

2. Because it is a proportional series.

So. a1(1-q^n)/(1-q)

a1=4 q=2

sn=2^2+n-5

1) sn=2an-1, the next term sn-1=(2an-1)-1sn-sn-1=an, gives an an-1=2 3;A2=3 is obtained from a1=2 and sn=2an-1; Finally, the proportional sequence an=3*(2 3) (n-2) is obtained

Send it to you by email, the answer here is not clear, especially the number.

The title of the mountain is too vague.

If a(n+1)=2 an +1, then:

a2=2/a1+1=2/1+1=3

a3=2/a2+1=2/3 +1=5/3

a4=2 a3+1=2 (5 3)+1=11 5a5=2 (11 reed5)+1=21 11 If a(n+1)=2 (an +1), then:

a2=2/(a1+1)=2/(1+1)=1a3=2/(a2+1)=2/(1+1)=1………The clan was in an uproar.

A sequence is a constant sequence of numbers where each item is 1.

a5=1

(1)a(n+1)=an/(2an+1)

1 a(n+1)=(2an+1) an=1 an +21 a(n+1)-1 an=2, which is a fixed value.

1 a1 = 1 1 = 1, the series is a series of equal differences with 1 as the first term and 2 as the tolerance 1 an=1+2(n-1)=2n-1

an=1/(2n-1)

When n=1, a1=1 (2 1-1)=1, and the general formula for the series of common terms is an=1 (2n-1).

2)bn=2ⁿ/[1/(2n-1)]=2ⁿ·(2n-1)tn=b1+b2+..bn=1×2+3×2²+5×2³+.2n-1)×2ⁿ

2tn=1×2²+3×2³+.2n-3)×2ⁿ+(2n-1)×2^(n+1)

tn-2tn=-tn=2+2²+.2ⁿ-(2n-1)×2^(n+1)

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

tn=(n-1)×2^(n+2) +2

a2 = a1 2 + 1 2, 2a3 = a2 + 2, 4a4 = 2a3 + 6, the first three formulas add 4a4 = a1 2+.

That is, the last two crosses are enlarged by 2 and 4 times respectively, and the three are added. I made a mistake at first, I'm sorry!

The title is a little unclear, Ming Shu digs! Is it 1 (2AN), or is it an 2, and I have done a few other things, but I still don't have an answer.

It's "an=2a(n-1)+2 n-1", right?

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

an - 1 = 2 [ a(n-1) -1 ] 2 n, divide by 2 n on both sides at the same time.

an - 1)/2^n = 2×[ a(n-1) -1 ]/2^n +1

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

an - 1)/2^n - a(n-1) -1 ]/2^(n-1) =1,n≥2

That is, the number series is (a1-1) 2=2 as the first term, and 1 is the equal difference number series of the common sock precession.

an - 1)/2^n=2+(n-1)×1=n+1

an - 1=(n+1)·2^n

Let the sum of the first n terms of the series be tn

then tn= 2 2 + 3 2 +4 2 +n+1)·2 n①

2tn= 2×2² +3×2³ +n·2^n + n+1)·2^(n+1) .Get. tn= (n+1)·2^(n+1) -2²+2³+…2^n) -4

n+1)·2 (n+1) -2+2 +2 good completion + ......2^n) -2

n+1)·2^(n+1) -2(1-2^n)/(1-2)] 2

n+1)·2^(n+1) -2^(n+1)

n·2^(n+1)

sn=tn + n=n·2^(n+1) +n

The title is too vaguely written.

If a(n+1)=2 an +1, then:

a2=2/a1+1=2/1+1=3

a3=2/a2+1=2/3 +1=5/3

a4=2/a3+1=2/(5/3)+1=11/5a5=2/(11/5)+1=21/11

If a(n+1)=2 an +1, then:

a2=2/(a1+1)=2/(1+1)=1a3=2/(a2+1)=2/(1+1)=1………

A sequence is a constant sequence of numbers where each item is 1.

a5=1

This question is not of much value. According to the problem a(n+1)=2 an+1, substitution a1=1 can get a2=3, a3=5 3, a4=11 5, a5=21 11

It's a periodic series of numbers, and we're going to write out a couple of things to analyze.

a2=2 a3=3 a4=1 a5=2 a6=3 a7=1………

So an=n is the remainder after dividing by 3.

2。Because 0 and note a4=a1+3 so a5=a4 3=a1 3+1 1a8=a7 3=a1 6+2 3 note 0 so this sequence is a series with a period of 7.

After that, let's push it yourself, and the general item doesn't seem to be so easy to write.

1. By definition, a1=1, a2=a1+1=2, a3=a2+1=3, a4=a3 3=1, a5=a4+1=2,.

As you can see, this is a series of periods with period 3, so an=n-[(n-1) 3]*3, where [x] denotes the largest integer that does not exceed x.

(1)an<3,a(n+1)=an+1,an=n

an>3,a(n+1)=an/3,an=(1/3)^(n-2)

Using the summation formula, you can write the second question. Do the math yourself and won't ask me again.