Given the equation x & sup2 / / A & sup2; + Y & sup2 / / B & sup2; = 1 (a > b > 0), point P (- 3,1) is on the straight line x = - A & sup2 / / C, the incident light passing through point P and direction vector a = (2, - 5) is reflected through the left focus of the ellipse after y = - 2, then the eccentricity of the ellipse is 0

Given the equation x & sup2 / / A & sup2; + Y & sup2 / / B & sup2; = 1 (a > b > 0), point P (- 3,1) is on the straight line x = - A & sup2 / / C, the incident light passing through point P and direction vector a = (2, - 5) is reflected through the left focus of the ellipse after y = - 2, then the eccentricity of the ellipse is 0

The idea is that the slope of the reflected ray and the incident ray are opposite to each other. Use the intersection point C of y = - 2 and the straight line passing P with a slope of (- 5 / 2) and the slope of C and the focus to know, and then use - A / C = - 3. I'll call you on my mobile phone

If the angle between two asymptotes of hyperbola is 60 ° then the eccentricity of the curve is 0

Let the eccentricity of the curve be e = C / a [C ^ 2 = a ^ 2 + B ^ 2] when the hyperbola is x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0), the two asymptotes of the hyperbola are y = + / - (B / a) x, so: + / - (B / a) = tan60 ° = root 3, so the eccentricity of the curve is e = C / a E = C / a e ^ 2 = C ^ 2 / A ^ 2 = (a ^ 2 + B ^ 2) / A ^ 2 = 1

Given the asymptote equation, how to get the hyperbolic equation
Example: given the asymptote of hyperbola 2x ± y = 0 and passing through point (1,3), we can solve hyperbolic equation
Let the hyperbola be 4x ^ 2-y ^ 2 = k, (1,3) be substituted to get k = - 5, and the hyperbolic equation be y ^ 2 / 5-4x ^ 2 / 5 = 1
There's a formula for that, right?
Well, draw inferences from one instance
1. If the asymptote equation is y = ± 1 / 3, what is the hyperbolic equation?
Is X & sup2 / / 9b & sup2; - Y & sup2 / / B & sup2; = ± 1

If the hyperbola is:
X ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = K (k is a constant, can be positive or negative) ---- - - (1)
(if k > 0, it can be reduced to: x ^ 2 / (a (root K) ^ 2-y ^ 2 / (b (root K)) ^ 2 = 1;
Such as K

Let AK = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 ', + K ^ 2 K belong to positive integer, then the sum of the first n terms of the sequence 3 / A1 5 / A2 7 / A3', (2n + 1) / an ',, is?

ak=1/6k(k+1)(2k+1)
Sn=3/[1/6*1*(1+1)*(2*1+1)]+5/[1/6*2*(2+1)*(2*2+1)]+.+(2n+1)/ [1/6n(n+1)(2n+1)]=6/[1*(1+1)]+6/[2*(2+1)]+.+)]+.+6/ [n(n+1)]=6(1-1/2)+6(1/2-1/3)+.+6(1/n-1/(n+1))=6(1-1/2+1/2-1/3+1/3-1/4+.+1/(n-1)-1/n+1/n-1/(n+1)]=6(1-1/(n+1))

The known sequence {an} is an arithmetic sequence, BK = a1 + A2 + a3 + +AK / K (k is a positive integer)
(1) Verification: sequence {BN} is also arithmetic sequence
(2) If A1 = 1 and (a1 + A2 +...) +A13）:(B1+B2+…… +B13) = 3:2, find the general formula of these two sequences {an} {BN}

(1) Let an = a1 + (n-1) * D, because BK = a1 + A2 + a3 + +In AK, a2-a1 = D = a3-a2 = D = ak-a (k-1), which satisfies the property of arithmetic sequence, so BK is also arithmetic sequence

(2 + 1 / N) ^ n limit

The limit of (1-1 / N) Λ 2 =?

（1-1/n)∧2=1-2/n+1/n^2
When the limit is taken, the latter two terms are both 0, and the original formula is equal to 1

N ^ 2 / (n ^ 2 + π) why is the limit 1?
Recently in looking at the postgraduate entrance examination topic, for a long time did not contact, encountered a lot of limit is 1 function, has been wondering why is 1, seek to wake up
Wrong, right
n^2/(n^2+πn）

lim(n-->∞） n^2/(n^2+π）
=lim(n-->∞） 1/(1+π/n^2）
=1

What is the limit of 1 + 2 + 3 +... + (n-1) / N ^ 2 when n →∞

[1 + 2 + 3 +... + (n-1)] / N ^ 2 = 2n (n-1) / N ^ 2 = (2n ^ 2-2n) / N ^ 2 = 2-2 / N, so the limit is 2

6 ^ (1 / 2 ^ n) + 6 ^ (1 / 2 ^ (n-1)) +... + 6 ^ 1 / 2 for limit

When n is a positive integer value, 1 / 2 ^ n > 0, so 6 ^ (1 / 2 ^ n) > 1
When n tends to infinity, the above formula is larger than the sum of N ones, so it diverges