Given sequence 1, A1, A2, a3,4 into arithmetic sequence, 1, B1, B2, b3,4 into proportional sequence, find (a1 + A2) / B2

Given sequence 1, A1, A2, a3,4 into arithmetic sequence, 1, B1, B2, b3,4 into proportional sequence, find (a1 + A2) / B2

Let the tolerance of sequence 1, A1, A2, A3 and 4 be d
4-1=4d
d=3/4
Let the common ratio of 1, B1, B2, B3 and 4 be Q
4/1=q⁴
q⁴=4
q²=2
(a1+a2)/b2
=(1+d+1+2d)/(1·q²)
=[1+ 3/4+1+2·(3/4)]/2
=17/8

Because we can't figure out the figure, we only have a simple description [the figure is roughly: a line segment AC has five points a, B, C, N and M. from left to right, the order is a, B, N, m and C. then, M is the midpoint of BC, and n is the midpoint of the whole line segment AC.]
Given the line AB = 8cm, find the length of M and n
This is five questions on page 65 of the learning guide. 】

MN=NC-MC=1/2AC-1/2BC=1/2(AC-BC)=1/2AB=4cm

Let each term of a sequence be the sum of the corresponding terms of an arithmetic sequence and an arithmetic sequence. If C1 = 2, C2 = 4, C3 = 7, C4 = 12, the general term formula cn is obtained

Let the arithmetic sequence a1 + (n-1) * D, the arithmetic sequence b1q ^ (n-1), cn be the sum of them, and then solve four unknowns,
a1+b1=2
a1+d+b1q=4
a1+2d+b1q^2=7
a1+3d+b1q^3=12
We get that A1 = B1 = D = 1, q = 2
So CN = n + 2 ^ (n-1)

Each term of the sequence {CN} is the sum of the corresponding terms of the arithmetic sequence and the proportional sequence with the first term of 1. If C3 = 7, C4 = 12, find the first n terms and Sn of the sequence {CN}

Let the arithmetic sequence be an and the proportional sequence be BN
So an = 1 + (n-1) d
Bn=q^(n-1)
So C3 = 1 + 2D + Q ^ 2 = 7
C4=1+3d+q^3=12
So q = 2, d = 1
So Sn = (1 + 2 + 3 +... + n) + (1 + 2 + 4 + 8 +... + 2 ^ n-1)
=n(n+1)/2+(1-2^n)/(1-2)
=2^n+(n^2+n)/2-1

It is known that SN is the sum of the first n terms of the sequence {an}, and 2Sn = 3an − 2 (n ∈ n *) (I) find the sum of an and Sn; (II) if BN = log3 (Sn + 1), find the sum of the first n terms of the sequence {b2n}

When n ≥ 2, 2sn-1 = 3an-1-2, 2sn-2sn-1 = 3an-3an-1, 2An = 3an-3an-1, an = 3an-1, an = 3an-1, an {an} is an equal ratio sequence with the first term of 2 and the common ratio of 3, an = 2 · 3n-1, Sn = 2 (1 − 3n) 1 − 3 = 3n-1

Let the first n terms and Sn of the sequence an, and Sn = 2an-2, n belong to positive integer, (1) find the general term formula of the sequence an, (2) let CN = n / an, find the first n terms and TN of the sequence
Let the first n terms of the sequence an and Sn, and Sn = 2an-2, n belong to positive integers,
(1) (2) let CN = n / an, find the first n terms and TN

1, S1 = A1 = 2a1-2, A1 = 2Sn = 2an-2 (1) s (n + 1) = 2A (n + 1) - 2 (2) (2) - (1): a (n + 1) = 2A (n + 1) - an a (n + 1) = 2An. Therefore, {an} is an equal ratio sequence with the first term of 2 and the common ratio of 2 )2,cn=n/2^nTn=1/2+2/2^2+3/2^3+… +n/2^n (3)(3)/2：Tn/2=1/...

In the positive integer sequence, the first n terms and Sn satisfy Sn = 1 / 8 * (an + 2) ^ 2, and it is proved to be an arithmetic sequence. If CN = 1 / (an * an + 1), find the first n terms and TN of CN

Sn = 1 / 8 * (an + 2) ^ 2 has sn-1 = 1 / 8 * (an-1 + 2) ^ 2 subtraction; an = 1 / 8 * (an + 2) ^ 2-1 / 8 * (an-1 + 2) ^ 2 arranges (an-an-1-4) * (an + an + 1) = 0 positive integer sequence an-an-1 = 4An arithmetic sequence Sn = 1 / 8 * (an + 2) ^ 2 has A1 = 1 / 8 * (a1 + 2) ^ 2A1 = 2An = 4n-2sn = 2n ^ 2cn = 1 / (an * an + 1) = 1 /

As shown in the figure, in △ ABC, AI and Bi bisect ∠ BAC and ∠ ABC respectively. CE is the bisector of ∠ ACD at the outer angle of △ ABC, intersecting Bi extension line at e, connecting CI. (1) when △ ABC changes, let ∠ BAC = 2 α. If α is used to represent ∠ BIC and ∠ E; (2) if AB = 1 and △ ABC is similar to △ ice, calculate the corresponding AC length

(1) In △ BCE, there are: ∠ e = 180 ° - ∠ BCE - ∠ CBE, and ∠ ECI is half of the flat angle ∠ BCD, ≠ ECI = 90 °, ≠: ∠ e = 90 ° - ∠ BCI - ∠ CBE. In △ ABC, 12 ∠ BAC = 12 (180 ° - ∠ ABC - ∠ ACB) = 90 °-- ∠ BCI - ∠ CBE, ≠ e = α (2) when △ ACB ∽ ice, ∠ ACB = ∠ ice = 90 °, ABC = ∠ IEC = α, so α = 30 ° AC = 12. (3) when △ BAC ∽ ice, ∠ BAC = ∠ ice = 90 °, IEC = 12 ∠ BAC = 45 °, so ∠ ABC = ∠ ACB = 45 ° AC = AB = 1

In △ ABC, Bi and CI are bisectors of ∠ ABC, ∠ ACB, BM and cm are bisectors of ∠ ABC and ∠ ACB, respectively

The conclusion should be ∠ BIC + ∠ M = 180 ° and the proof is as follows. By examining several angles with B as the vertex, ∵ Bi bisection ∠ ABC, ∵ IBC = ∠ ABC / 2, ∵ BM is the bisection line of the outer angle of ∠ ABC, ∵ CBM = (1 / 2) (180 ° - ABC) = 90 ° - ABC / 2, ∵ IBM = ∠ IBC + ∠ CBM = ∠ ABC / 2 + (90 ° - ABC / 2) = 90 °

In △ ABC, Bi and CI divide ∠ ABC and ∠ ACB equally, and ∠ a = ∠ α, then calculate the degree of ∠ BIC

In △ BCI, ∠ BIC = 180 ° - (∠ IBC + ∠ ICB) = 180 ° - 12 (180 ° - a), i.e. ∠ BIC = 90 ° + 12 ∠ a