Urgent! to find the general term of the number series 2 30

Question 1: The first step is to find a1, s1=a1=(a1+1) 2 4, and solve a1=1;

The second step consists of sn=(an+1) 2 4 (1).

s(n-1)=(a(n-1)+1) 2 4 (2).

From (1)-2), 4an=(an+1) 2-(a(n-1)+1) 2, collated.

An 2-a(n-1) 2=2(an+a(n-1)), because an>0, so both sides of the equation are divided by an+a(n-1), an-a(n-1)=2, obviously an is the first difference series with a tolerance of 1 and 2, an=1+(n-1)*2

Question 2: The first step is to find a1, s1=a1=4a1+2, a1=-2 3;

The second step is composed of sn=4an+2 (1).

This yields s(n-1)=4a(n-1)+2 (2).

Obtained from (1)-2), an=4an-4a(n-1), collated.

an a(n-1) = 4 3, obviously an is a proportional series with -2 3 as the first term and 4 3 as the common ratio. You write the general term formula yourself.

The third question is the same as the second question, it is also a proportional series, and the method is the same as the second question.

The fourth question is slightly different, so I need to use the formula an=sn-s(n-1), and I'll write it again (it's troublesome to write).

an+2sns(n-1)=0, which is obtained by using the formula an=sn-s(n-1).

sn-s(n-1)+2sns(n-1)=0, divided by sns(n-1) on both sides.

1 s(n-1)-1 sn+2=0, finishing 1 sn-1 s(n-1)=2, s1=a1=1 2, 1 s1=2; It can be seen that 1 sn is actually a series of equal differences with the first term 2 and the tolerance is 2, 1 sn=2+(n-1)*2=2n, so sn=1 2n, an=sn-s(n-1), you can find an, because I wrote more, there may be mistakes, please forgive me for the mistakes.

Divide the items of the series a1=(1+1) 2

a2=(2+0)/2

a3=(3+1)/2

a4=(4+0)/2...

an=1 2=[2n+1-(-1)npower] 4 The process should be quite detailed.

a1=2a2=a1+1

a3=a2+2

a4=a3+3

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

The two sides add Changyin Circle, get:

a1+a2+..an=a1+a2+..a(n-1)+[2+1+2+..n-1)]

Eliminate the same caution on both sides, and get:

an=1+2+3+..n-1)+2

n(n+1) collapse resistance 2+1

a1=1a2=a1+1

a3=a2+2

a4=a3+3

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

The brothers on both sides celebrate Liang Jia, and they have a bad trace: envy luck.

a1+a2+..an=a1+a2+..a(n-1)+[1+1+2+..n-1)]

Remove the same term on both sides to get:

an=1+2+3+..n-1)+1

n(n-1)/2+1

The first question is to use the method of undetermined coefficients.

We set. an+x)=2(a(n-1)+x)an+x=2a(n-1)+2x

an=2a(n-1)+x

Get x=1 and set bn=an+1

then bn=2b(n-1).

b1=1+1=2

Then it is a proportional series with the first term being 2 and the common ratio being 2.

Then bn=2 n, i.e. an+1=2 n

Rule. an=2^n-1

The second question is to accumulate Wang's early summoning method.

an-a(n-1)=n

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

a(n-2)-a(n-3)=(n-2)

Open your mouth. a2-a1=2

a1=1an=1+2+3+

Needless to say, the rest will be said.

Hello landlord: 1: an=2a(n-1)+1

Add "1" on both sides of the equation, that is, an+1=2a+2, which is an equal ratio of the front jujube column, so an+1=(a1+1)2 (n-1) so an=2

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

a(n-2)-a(n-3)=n-2

a3-a2=3

a2-a1=2

Add the above formulas to the nuclear base index, so an=2+3+4+......n,an=(n +n-2) 2 Thanks! o( Reconfiguration )o

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

This is a series of proportional numbers.

The first item is 2, and the ratio is 2

The formula is Bu Min slip an=2n power-1

The second one I'll never think about.

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

an+1 is to talk about a proportional imitation or sequence of cherry blossoms in the common ratio of 2.

Then it's good to beg.

The second is superimposed, and the reserve is an-1=1+2+.n and then equal the sum of the difference series, it is too simple.

To distinguish between them, the sequences are capitalized a

an+1=an +2n+1

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

a2=a1+2*1+1

The state of the upper beam surface is quietly added up in several forms, and the sail slag is obtained.

an+1 -a1=2(n+n-1+..1)+nan+1=(n+2)n-1

an=n^2-2

There is a recursive formula: a(n+1)-an=n

i.e. a2-a1=1

a3-a2=2

.an-a(n-1)=n-1

The above n-1 equations are added to give :

an-a1=1+2+..n-1

an-1=n(n-1)/2

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

Removed the minus sign, which is an=6n-5

So, the original an=(-1) n*(6n-5).

(6n+1)×（1）^n.Don't look at the symbols are equal difference sequences, and then add signs according to the symbols.