The number series problem is urgently needed, the number series problem, urgent

an=sn-s(n-1)=2 n-2 (n-1)=2 (n-1).

Then the terms are then squared and an=2 (2n-2), which is an=4 (n-1).

According to the general formula of the proportional series, a1=1, q=4

Then according to the sum of the first n terms of the proportional series, sn=(4 n-1) 3 is obtained

an=sn-s(n-1)=2 n-2 (n-1)=2 (n-1).

So. a1^2+a2^2+……an^2

0+(2^1)^2+(2^2)^2+..2^(n-1))^22^2+2^4+2^6+..2^[2(n-1)]4+4*4+4*4^2+..

4*4 (n-2) is the sum of the proportional series with the first term is 4 and the common ratio is 4, a total of n-1 terms.

Original = 4(1-4 (n-1)) (1-4)4(4 (n-1)-1)./3

Isn't it like this: sn=2 (n-1)-(a1) 2 ??

If this is the case, there is sn+1=2 n-(a1) 2an+1=sn+1-sn=2 (n-1).

So there is: a2 = 1, a3 = 2

a2^2+……an^2=2^0+2^1+2^2+..2^(n-1)=(2^n ) 1

In a proportional series, it is known that for any positive integer n, a1+a2+......an=(2 n)-1, then a1 2+a2 2+......An 2 is equal to a [ 2 n)-1] 2 b 1 3 [(2 n)-1] c 1 3[(4 n)-1] d 4 n-1

Let's see if it's true

If you don't understand it, write it clearly, I will definitely help you solve it, you can send me a message, and we will study it again.

a1=s1=1

When n>=2, an=sn-s(n-1)=2 n-2 (n-1)=2 (n-1).

So a1 2+a2 2+......an^2

1+(2^1)^2+(2^2)^2+..2^(n-1))^21+2^2+2^4+2^6+..2^[2(n-1)]1+4+4^2+4^3+..

4 (n-1) is the sum of the proportional series with the first term 1 and the common ratio of 4, a total of n terms:

Original formula = 1*(1-4 n) (1-4) = (4 n-1) 3

It seems to be a problem in high school, and it's a little bit difficult to see the known conditions.

I was only in elementary school, so I didn't go to school

I'm sorry, can I be clearer, I don't understand.

an=[under the root number (an-1)+1] square an 0 under the root number an=under the root number (an-1)+1 let bn= under the root number an b1=under the root number 2

bn = b(n-1) + 1 bn = 2 + n-1 an = (2 + n-1) squared under the root number.

5(a1+a1q)=4(s1+a1q+a1q^2+a1q^3) 4q^3+4q^2-q-1=0 (q+1)(4q^2-1)=0

Proportional series q=1 2

sn=2a1( bn=1 2 +2a1( If the proportional sequence a1=-1 4

an=sn-s(n-1)=1 Ziqiao2*(1 Shout 2) (n-1)-(1 2) (n-1)=-1 2*(1 2) (n-1).

So: a1=-1 2

and s1=(1 2) 1+a=1 2+a

s1=a11 2+a=-1 2

So: a=-1

2sn-bn+1+b1=0, get b1=-1 2

2sn-1 -bn-1 +1+b1=0

Subtract 2 formulas, and get 2sn-2sn-1 -bn+bn-1=0, 2bn-bn+bn-1 =0, bn=-bn-1, and the argument is that Zheng takes the stove with bn=(-1 2)*(1) (n-1).

Because of an 0, if a(n+1)=-(an+2), then a(n+1) is a negative number, which contradicts an 0, so it can only be a(n+1)=an+2

an=a1*q^(n-1)

an=ka[（n+1）+（n+2）+.q (n-1) infinite difference series (n+1) + (n+2) +The partial sum of the parts is sn=n(3n+1) 2, and the limit of sn is 3 2

So an=ka*3 2=q (n-1), k=q (n-1) (3 2a) don't know the range of a.