As shown in the figure, it is known that M is the midpoint of AB and N is the midpoint of AC. if Mn = 5cm, then BC=______  cm．

As shown in the figure, it is known that M is the midpoint of AB and N is the midpoint of AC. if Mn = 5cm, then BC=______  cm．

∵ m is the midpoint of AB, n is the midpoint of AC, ∵ an = Mn + MC, ① MB = an + Mn, ②, ① - ② get: mb-mc = 2Mn = 10, from the figure we can see that mb-mc = BC, ∵ BC = 10

As shown in the figure, given that the segment AB of point C is 5:3, the segment AB of point D is 3:5, and the length of CD is 10cm, what is the length of AB? (formula)

Let AC = 5x, BC = 3x, if my drawing is correct, then ad = CB = 3x (x is a multiple, when 3x represents the length)
And DB = 5x, so DC = 2x
2x=10
x=5
AB = 8x, so AB is 40

Given the line segment AB = 2cm, extend AB to C so that BC = 2Ab. If D is the midpoint of AB, then the length of line segment DC is______ ．

As shown in the figure: ∵ line AB = 2cm, BC = 2Ab, ∵ BC = 4cm, ∵ D is the midpoint of AB, ∵ BD = 1cm, ∵ DC = BD + BC = 1 + 4 = 5cm

In the equal ratio sequence {an}, it is known that for n belongs to N +, a1 + A2 +... + an = 2 ^ n-1, find A1 ^ 2 + A2 ^ 2 +... + an ^ 2

a1+a2+...+an=2^n-1=Sn
a1+a2+...+an-1=2^（n-1)-1=Sn-1
Sn-sn-1 = an = 2 ^ n, let an ^ 2 = BN = 2 ^ (n + 2), B (n-1) = a (n-1) ^ 2 = 2 ^ 2 (n + 1),
Obviously, BN = 2B (n-1), B1 = A1 ^ 2 = 4, SBN = 4 (2 ^ n-1) = 2 ^ n + 2-4

Let {an} be an example, TN = Na1 + (n-1) A2 +2an-1 + an, known T1 = 1, T2 = 4, (1) find the first term and common ratio of sequence {an}; (2) find the general term formula of sequence {TN}

(1) Let {an} take ratio as Q, then T1 = A1, T2 = 2A1 + A2 = A1 (2 + Q). ∵ T1 = 1, T2 = 4, ∵ A1 = 1, q = 2. (2) let Sn = a1 + A2 + +It is known from (1) that an = 2N-1. Sn = 1 + 2 + +2n-1=2n-1∴Tn=na1+（n-1）a2+… +2an-1+an=a1+（a1+a2）+… +（a1+a...

The sequence an satisfies: A1 = 1, A2 = 2 / 3, an + 2 = 3 / 2An + 1-1 / 2An, DN = an + 1-an, proving that DN is an equal ratio sequence
(2) The general term formula of the sequence an
(3) Let BN = 3n-2, find the first n terms and Sn of sequence an * BN

(1) A (n + 2) = 3 / 2A (n + 1) - 1 / 2 ana (n + 2) - A (n + 1) = 1 / 2 a (n + 1) - 1 / 2 and (n + 1) = (1 / 2) * DND (n + 1) / dN = 1 / 2, so {DN} is an equal ratio sequence, q = 1 / 2, the first term is 1 / 2 (2) a (n + 1) - an = (1 / 2) ^ n an-a (n-1) = (1 / 2) ^ (n-1), n ≥ 2... A2-a1 = 1 / 2 accumulation

In {an}, A2 = - 3, A5 = 36, then the value of A8 is ()
A. - 432B. 432C. - 216d. None of the above is true

According to the property of the equal ratio sequence, we can know that: a5a2 = a8a5 = Q3, then A8 = a52a2 = 362 − 3 = - 432

In the equal ratio sequence {an}, if A2 * A8 = 18, the value of A5 is equal to

a2*a8
=a5/q^3*a5*q^3
=a5^2
a5=18^1/2=±3√2

It is known that the common ratio of {an} is q = 3, the first three terms and S3 = 13 / 3
Find: (1) the general formula of {an}; (2) if f (x) = asin (2x + b) (a > 0, O)

a1+3a1+9a1=13 a1=1
an=a1q^(n-1)=3^(n-1)
a3=3^2=9
(2) If f (x) = asin (2x + b) (a > 0, O

It is known that the common ratio of an is 3, and the first three terms and S3 = 13 / 3 (1) are used to find an

Let the first term be A1, A2 = 3A1, A3 = 9a1, then S3 = a1 + A2 + a3 = a1 + 3A1 + 9a1 = 13a1, so A1 = 1 / 3, an = (1 / 3) * 3 ^ (n-1)