b1 6, b2 4, b3 3, bn 1 bn is the formula for finding the general term of the bun in the series of eq

Solution: Let an=bn+1-bn be a proportional series.

a1=b2-b1=-2 a2=b3-b2=-1 common ratio q=1 2 so an=bn+1-bn=-1 2 (n-2).

b2-b1=-1/2^(-1)

b3-b2=-1/2^0

b4-b3=-1/2^1

b5-b4=-1/2^2

bn-bn-1=-1 2 (n-2) The sum of all forms yields: bn-b1=-(2+1+1 2+1 4+......1/2^(n-3)

4+1/2^(n-3)

So bn=2+1 2 (n-3).

Let the common ratio be q, and bn=q (n-1)b2+b3=12

q+q^2=12

q^2+q-12=0

q+4)(q-3)=0

q = -4 or as in 3

So bn=(-4) (n-1), or bn=3 (n-1).

Solution: 1. Suppose the general term bn = b1*q (n-1) of the proportional series.

From the characteristics of the proportional series, it can be seen that b1 and q are not equal to 0.

2. B5=162 is known in the question, and the general term is substituted.

Available: 162=b1*q 4

3. 3b2+2b3=b4 is known in the question, and the general term is substituted.

It can be obtained: 3b1*q + 2b1*q 2=b1*q 3 equal signs on both sides of the same divide by b1*q.

3+2q=q^2

The upper group is a typical one-dimensional quadratic equation.

4. Solve the equation q 2-2q-3=0

Using the cross multiplication, q+1)(q-3)=0 gives q=-1 or 3

5. Substitute Q and discuss.

1) When Q=-1, B1 (1) 4=162, we can know that B1=162 and Bn =162 (1) (N-1).

2) When q=3, b1 3 4=162, it can be seen that b1=2, bn=2 hu or sleepy 3 (n-1).

3b₁q＋2b₁q²＝b₁q³，①

b₁q⁴＝162，②

Solution b 162, b 1;

or B2, Q3, so bn 162 (1) sn 81[1 (1) ].

or bn split-wheel leaky tung stool 2 3 sn 3 1 . Wanton.

1) Qin imitation B1 + B2 = 30, B3 + B4 = 120, find B5 + B6B3 + B4 = B1Q 2 + B2Q 2 = (B1 + B2) Q 2 = 120 So: the first grinding bucket Q 2 = 4

b5+b6=b3q^2+b4a^2=(b3+b4)q^2=120x4=480

b1b2b3=1 slag 8, b1+b2+b3=21 8 These two shirts can be found quietly.

b1 = 1 8, common ratio = collapse 8 = 2 (2n-5).

an=log2bn=2n-5

Summary. In the proportional series bn + b1b3 = 3b2, b5 = 27b2 + find the first n terms and sn

Wait a minute, Takako.

Baozi, are you sure this is a series of equal differences?

Can you take a picture of the topic and send it to me?

Okay, I see that.

I was replying to someone just now, so you have been waiting for a long time.

Right away. Sorry for keeping Poko waiting.

The answer is: sn=69 49n-39 49n(n-1) This question looks weird.

It's not an integer.

Please take a look at it.

So, the answer is: sn=69 49n-39 49n(n-1) I hope it will help you.

I also hope that Baozi will give the teacher a feedback after seeing the answer.

Okay. I hope to ask me more questions in the future.

Ask about custom messages].

bn=2 (n-1)b(n+2)-b(n+1)=2(b(n+1)-bn), and then multiply (b(n+2)-b(n+1)) b2-b1)=2 n, i.e., b(n+2)-b(n+1)=2 Accumulate the line of songs, and the ridge is bn-b1=2 (n-2)+2^0bn=2^(n-2)+.2^0+2^02bn=2^(n-1)+.

2 1 + 2 1 minus bn = 2 (n-1) + 2 1-2 0-2 0 = 2....

Because it is a series of equal differences, let the equal difference number be n, then b2=b1+n=2+n, b3=b2+n=2+2n, b4=b3+n=2+3n. So b1 + b2 + b3 + b4 = 8 + 6n = 26, and the winner bends n = 3. So bn=2+3(n-1) is a sign, n>0 and an integer.

Let dn=b3n-2=2+3(3n-2-1)=2+9(n-1), so u10 is the first 10 terms of dn and the digging rent, d1=2, d10=2+9(10-1)=83, so u10=10(d1+d10) 2=