High school math, solving, process, thanks! 10

Solution: a1=3, an+1=2an+3

an+1+3=2(an+3), a1+3=6, the series is a proportional series with 6 as the first term and 2 as the common ratio, an+3=6 2n-1=3 2n, an=3 2n 3=3(2n-1), sn=3[(21-1)+(22-1)+(23-1)+....2n-1）]=3[ 2⎛ 1-2n1-2-n]=3(2n+1-2-n)．

So the answer is: 3(2n+1-2-n) solution: a1=3,an+1=2an+3

an+1+3=2(an+3), a1+3=6, the series is a proportional series with 6 as the first term and 2 as the common ratio, an+3=6 2n-1=3 2n, an=3 2n 3=3(2n-1), sn=3[(21-1)+(22-1)+(23-1)+....2n-1）]=3[ 2⎛ 1-2n1-2-n]=3(2n+1-2-n)．

So the answer is: 3(2n+1-2-n).

Count a few more items. a1=3

a2=2*3+7=13

a3=2*13+7=33

a4=2*33+7=73

a5=2*73+7=153

It was found that a(n+1)-an is a proportional sequence with a prime minister of 10 and a common ratio of 2.

So a(n+1)-an=10*2 (n-1) because a(n+1)=2an+7.

a(n+1)-an=an+7=10*2 (n-1), so an=10*2 (n-1)-7

Answer: ABCOS 2(c 2) = 1

ab*(1+cosc)/2=1

ab+abcosc=2

c²=a²+b²-2abcosc

4=a²+b²-2(2-ab)

8=a²+b²+2ab

i.e. (a+b) =8

a+b=2√2

i.e. the sum of the distance from c to a and the distance from c to b is the constant 2 2 (greater than 2) using the definition of an ellipse, the trajectory of c is an ellipse.

2a=2√2， 2c=2

a=√2，c=1

b =1 The elliptic equation is x 2 + y = 1 (a, b on the x-axis) or The elliptic equation is y 2+x =1 (a, b on the y-axis).

Question f(x) 3sin(2x+6) minimum positive period, axis of symmetry, center of symmetry.

The minimum positive period is 2 W

w is the coefficient before x.

The axis of symmetry is to make the brackets equal to zero, and then the resulting x + 2k center of symmetry is to make the k of the axis of symmetry equal to 0

The tolerance of the series of equal differences is known, let the sum of the precedes be , and find ;

) to find the value of such thatAnswer( )

16, I can't remember, I haven't read a math book for a long time.

1. The two straight lines are parallel, that is, the slopes are equal. Let the straight line be 4x+y+m=0 and substitute the point a(3,2) into 4x+y+m=0 to get 12+2+m=0, that is, m=-14

The equation for a straight line is 4x+y-14=0

2. Find the straight line in two-point formula.

3. The slope of the two straight lines is equal, that is, kk1=-1

Let the straight line be y=kx+b

From the straight line 2x+y-5=0, that is, y=-2x+5, its slope k1=-2 is obtained because it is perpendicular to its slope, so the slope of the straight line k=-1 k1=-1 (-2)=1 2

Substituting k=1 2 into y=kx+b, getting y=1 2x+b, and then substituting the point b(3,0) to obtain, 3 2+b=0, that is, b=-3 2, that is, the equation is y=1 2x-3 2, x-2y-3=0

Let y=-4x+b, then 2=-4*3+b, b=14. So the equation is y=-4x+14

kmn=(2+5) (1+1)=7 2, so let y=7 2x+b, and substitute (2,-3) to obtain: b=-10, so the equation is y=7 2x-10

k=1 2, let y=1 2x+b, substitute (3,0) to obtain, b=-3 2, so the equation is y=1 2x-3 2

1 Let the straight line be y kx+b, because the two straight lines are parallel so the slope is equal k1=k2=-4, and the gradient of the point a 3,2 and the slope of the equation 3=-4 2+b b=11 y=-4x+11 2 Let the straight line over m n be y=kx+b and substitute these two points into the equation to get k=7 2,b=-3 2 The equation for m n is 2y-7x+3=0, and the equation is equal to the equation y=kx+b for the point c(2,-3) The slope is equal to k=7 2 Substitute c,k into the equation to get b=-10 y=7 2x-10.