Let the left and right focus of hyperbola x2a2-y2b2 = 1 (a ＞ 0, B ＞ 0) be F1 and F2 respectively, and the right branch of the straight line intersection hyperbola passing through point F2 is at two different points m and N. if △ mnf1 is an equilateral triangle, then the eccentricity of the hyperbola is () A. 6B. 3C. 2D. 33

Let the left and right focus of hyperbola x2a2-y2b2 = 1 (a ＞ 0, B ＞ 0) be F1 and F2 respectively, and the right branch of the straight line intersection hyperbola passing through point F2 is at two different points m and N. if △ mnf1 is an equilateral triangle, then the eccentricity of the hyperbola is () A. 6B. 3C. 2D. 33

According to the meaning of the title, m and N are symmetric about x-axis, | NF2 | = B2A, | F1F2 | = 2c, ∵ △ mnf1 are equilateral triangles. Combined with the definition of hyperbola, we get MF1 = MF2 + 2a, ∵ (B2A + a × 2) × 32 = 2c, ∵ 3 (C2 + A2) = 4ac, divide both sides by A2, we get 3e2 − 4E + 3 = 0, and the solution is e = 3 or E = 33 < 1 (rounding off); so we choose B

The two focuses of hyperbola x square / N - y square = 1 (n > 0) are F1 and F2. P satisfies Pf1 + PF2 on hyperbola
If the two focuses of hyperbola x square / N - y square = 1 (n > 0) are F1 and F2, and P satisfies Pf1 + PF2 = 2 radical (n + 2) on hyperbola, then the area of triangle pf1f2 is?

Let Pf1 = P, PF2 = qp-q = 2A = 2 √ NP + q = 2 √ (n + 2) (P + Q) ^ 2 - (P-Q) ^ 2 = 4pq = 8pq = 2f1f2 = 2C = 2 √ (n + 1) by cosine theorem cosf1pf2 = (P ^ 2 + Q ^ 2-f1f2 ^ 2) / 2pqp ^ 2 + Q ^ 2 = (P-Q) ^ 2 + 2pq = 4N + 4, so cosf1pf2 = (4N + 4-4n-4) / 4 =

(a + b) = m, (a-b) = n, the cube of (AB) is expressed by the formula containing m and n

[(m^2-n^2)^3]/64

Given a ^ 2 = m, a ^ 3 = n, find a ^ 5 and a ^ 10 (expressed by the formula containing m, n)

2n
2n+1

It is known that the sum of the first n terms of the sequence (an) is Sn, and 2Sn = 3an-1. N belongs to the formula for finding the general term of n *,

∵2Sn=3an-1
∴2(Sn-Sn-1)=2an=3an-3a(n-1),(n≥2)
Let an = 3A (n-1) (n ≥ 2)
And 2S1 = 2A1 = 3a1-1, A1 = 1
According to the general term formula of equal ratio sequence, an = 3 ^ (n-1)

If the sum of the first n terms of an is Sn, and an = 2sn-3, then the general formula of an is_______

an=2sn-3
a(n-1)=2s(n-1)-3 (n≥2)
an-a(n-1)=2an
an=-a(n-1)
q=-1
a1=2s1-3
s1=a1=3
an=3*(-1)^(n-1)

Let the first n terms of sequence {an} and Sn = 3an-2 (n = 1, 2,...) (I) prove that the sequence {an} is an equal ratio sequence; (II) if BN + 1 = an + BN (n = 1, 2,...) And B1 = - 3, find the first n terms and TN of the sequence {BN}

(1) syndrome: because Sn = 3an-2 (n = 1, 2,...) ，Sn-1=3an-1-2（n=2，3，…） So when n ≥ 2, an = sn-sn-1 = 3an-3an-1, we get an = 32An − 1

The sum of the first n terms of an is Sn, Sn = 1 + 3an
Why s (n) - S (n-1) = a (n) = 3 (a (n) - A (n-1))?

In other words, A1 = 1.3A = 1
n> 1, an = sn-sn-1 = 1 + 3an - (1 + 3A (n-1)) = 3an-3a (n-1),
That is, an = 3 / 2A (n-1), that is, an = - 1 / 2 * (3 / 2) ^ (n-1)

The sum of the first n terms of the sequence {an} is Sn, and 2Sn + 1 = 3an, find an and Sn

When n = 1, 2S1 + 1 = 3A1, that is, A1 = 1, because 2Sn + 1 = 3an, so 2Sn + 1 + 1 = 3an + 1
An + 1 = 3an is an equal ratio sequence, and the common ratio is 3, so an = 3 ^ (n-1)
sn=3^n-1/2.

It is known that the sum of the first n terms of the sequence {an} is SN. If an = 2n (n + 2), then S10 = ()
A. 175132B. 1112C. 116D. 17566

∵an=2n(n+2)=1n−1n+2，∴Sn=(1−13)+(12−14)+(13−15)+… +(1n − 1 − 1n + 1) + (1n − 1n + 2) = 1 + 12 − 1n + 1 − 1n + 2. Then S10 = 32-111 − 112 = 175132