Proportional series a1 2 a2 2 a3 2 ,,,,,,,,,,, an 2 4n 1 3, find the value of a1 a2 a3 a4,,,,,,,,,,

If the common ratio of the proportional sequence is expressed by q, then the sequence is still the proportional sequence, but the first term becomes a1 2, the common ratio becomes q 2, and you can calculate the rest yourself.

a(n)=aq^(n-1)

a(n)]^2=a^2*(q^2)^(n-1),a(1)]^2=(4-1)/3=1=a.

If q=1, then, (4 n-1) 3=na 2=nContradiction.

Therefore, q is not equal to 1

4^n-1)/3=a^2[(q^2)^n-1]/(q^2-1)=[q^(2n)-1]/(q^2-1),4^2-1)/3=[q^4-1]/(q^2-1)=q^2+1,q^2=4

q=2, a(1)+a(2)+a(n)=[q^n-1]/[q-1]=2^n-1

q=2, a(1)+a(2)+a(n)=[1-q^n]/[1-q]=[1-(-2)^n]/3

a1 2+a2 2+ a(n-1) 2=(4n-5) 3 is 2 formula, the original formula is 1 formula, 1 formula minus 2 formula, an 2 = 4 3 where n is not equal to 1, when n is 1, a1 2 = 1, and an is an proportional series; When n is not 1, the common ratio is + or -1, because an 2 = 4 3, so a2 is 4 3 under the root number, but a1 is + or -1, a2 a1 is not + or -1, that is, an is not a proportional series, which contradicts the original question, so there is a problem with this problem... Did you make a mistake? I've converted the input method several times.,I'm tired to death.。。。

Give it a hard share...

Solution: Let the first term of the proportional series be a, and the common ratio be q

Then the series is also proportional series, the first term is a, and the common ratio is q a1 + a2 + a3 + a4 + a5 = 3

a（1-q^5)/(1-q)=3

i.e. a(1-q 5) = 3(1-q) ......1）

a1^2+a2^2+a3^2+a4^2+a5^2=12a²（1-q^10)/(1-q²)=12

That is, a (1-q 10) = 12 (1-q ) (2) is obtained from (1) (2): 4 (1 + q) = a (1 + q 5) a1 - a2 + a3-a4 + a5 = a (1 + q 5) (1 + q) = 4

Using the first and second equations, find a1 and q (common ratio). Then you can ask for the answer you want.

Solution: Let the common ratio be q

Then q = a4 a1 = -4 (1 2) = -8, so q = -2

Then an=a1q (n-1)=1 2 (-2) (n-1)=-(-2) (n-2).

So |an|=2^(n-2)

a1|+|a2|+|a3|+.an|=2^(-1)+2^0+2^1+2^2+……2^(n-2)=1/2×(1-2^n)/(1-2)

1/2×(2^n-1)

2^(n-1)-1/2

Happy learning.

o(∩_o~

First, calculate the common ratio: q=-2

Because it is an absolute value, it is then converted into the sum of the first n terms of the proportional series of a1=1 2 and q=2.

So |a1|+|a2|+|a3|+.an|=1/2*（1-2^n）*2=1-2^n

a1+a3=-5

a1*a3=4

Using Veda's theorem, it is equivalent to solving an equation.

t^2+5t+4=0

t1=-1,t2=-4

a1=-1,a3=-4

ora1=-4,a3=-1

Just count it on a scale of 2.

The general formula found in the first step. From a2=2, a5=1 4, the ratio q=1 2, a1=4 is obtained.

The second step is to analyze the characteristics of the sum. a2a3a4/a1a2a3=a4/a1=q3，a3a4a5/a2a3a4=a5/a2=q3，..

Therefore, it can be deduced that this is a proportional series with a1a2a3 as the initial value, and the general formula is obtained by finding the value of b1=a1a2a3 and the value of its ratio q3.

The third step is the sum of items that can be calculated according to the summation formula of the proportional series.

Let the ratio of the number of slag in the original equal ratio series q q, then the ratio of the slag beam in the number series is -qa1 + a2 + a3 + a4 + a5 = 3

a1(1-q^5)/(1-q)··a

a1²+a2²+…a5²=12，a1^2(1-q^10)/(1-q^2)··b

a1-a2+a3-a4+a5=?

a1(1=q^5)/(1+q)··c

b/a=c=4

a1+a2+..an=2^n-1

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

an^2=4^(n-1)

An 2 is a proportional sequence in which 1 is the first term and 4 is the common ratio.

a1^2+a2^2+..an^2=(4^n-1)/3

Solution: Let the common ratio be q

a5=a2 *q 3 solves q=1 2, so a1=4 an=4* (1 2) (n-1) so that bn=an*a(n+1)= 8* (1 4) (n-1), so {bn} is the first n term of the proportional sequence with the first term 8 and the common ratio of 1 4, and the first n term of the original formula = bn and sn=8*[1-(1 4) n] (1- 1 4) =32 3 (1- 1 4 n).

You switch you, not one won't.