Let the sum of the first n terms of the sequence {an} be Sn, if an + 1 = 3Sn, A1 = 1, then an + 1 = 1=

Let the sum of the first n terms of the sequence {an} be Sn, if an + 1 = 3Sn, A1 = 1, then an + 1 = 1=

an+1=3Sn
an=3Sn-1
an=Sn-Sn-1=an+1/3-an/3
2an/3=an+1/3
q=an+1/an=2
a1=1
an=a12^(n-1)=2^(n-1)

What are the coefficients and degrees of each of the following polynomials?
（1）-3x+5x^2-x^3+1
(2)2xy^2+x^2y-x^3y^3-7

The coefficient is - 1,5, - 3, and the constant term is 1
2. For the sixth order quaternion of XY, the coefficient is - 1,2,1, and the constant term is - 7

There are two barrels of oil. The second barrel is 1.5 times of the first barrel. If 2.5kg is taken out of the second barrel and put into the first barrel, the two barrels are equal in weight
What's the weight of the first barrel of oil?

If the weight of the first barrel is XKG, the weight of the second barrel is 1.5xkg
1.5X-2.5=X+2.5
1.5X-X=2.5+2.5
0.5X=5
X=10

Derivative problem: F (x) = (LN (x-1) / (x + 1)), # 160; find the value of 2009 order derivative of this function at x = 0, that is, find f ^ (20
Derivative problem: F (x) = (LN (x-1) / (x + 1)), # 160; find the value of 2009 derivative of this function at x = 0, that is, find f ^ (2009) (0)

f(x)=Ln(x-1)/(x+1)=ln(1-x)-ln(1+x)
f'(x)=1/(x-1)-1/(1+x)
f''(X)=-1/(x-1)^2+1/(x+1)^2
f'''(x)=2/(x-1)^3-2/(x+1)^3
It can be concluded that
f^(n)(x)=[(-1)^(n-1)](n-1)!/(x-1)^n+[(-1)^n](n-1)!/(x+1)^n
So f ^ (2009) (0) = (- 1) ^ 2008 (2008!) / (- 1) ^ 2009 + (- 1) ^ 2009 (2008!)
=-2008!-2008!=-2*2008!.

（x-3）/（0.5）-（x+4）/（0.2）=1.6

（x-3）/（0.5）-（x+4）/（0.2）=1.6
2(x-3)-5(x+4)=1.6
2x-6-5x-20=1.6
-3x=27.6
x=-9.2

There is a cylindrical drainage pipe by the river with an inner radius of 0.8 meters and a water flow velocity of 2.5 meters per second. How many cubic meters of water can be drained from this pipe at most per minute

3.14×0.8×0.8×（2.5×60）
=3.14×0.64×150
=3.14×96
=301.44 M3

Let f (x) = 2x + 3x-2, then when x → 0 ()
A. F (x) is the equivalent infinitesimal of X. B. f (x) is the same order but not equivalent infinitesimal of X. C. f (x) is lower than x. D. f (x) is higher than x

The problem is transformed into finding the limit: limx → 02x + 3x − 2x, then there is: limx → 02x + 3x − 2x = limx → 02xln2 + 3xln31 = LN2 + Ln3, because the answer is not equal to 1, so it is not equivalent infinitesimal, only infinitesimal of the same order, so choose: B

6 / 9 - 0 = 1 / 3 + 0.25 = 3 / 5 - 0.3 = 4 / 5 + 0.7 = 7 / 11 - 2 = 1 / 6 - 5 / 6=
6 / 9 - 0=
One third + 0.25=
Three fifths - 0.3=
4 / 5 + 0.7=
7 3 / 11 - 2=
1-1 / 6-5 / 6=

Six out of nine - 0 = 2 / 3
One third + 0.25 = 7 / 12
Three fifths - 0.3 = 0.3
4 / 5 + 0.7 = 3.5
7 three out of eleven - 2 = 5 three out of eleven
1-1 / 6-5 / 6 = 0

Answer to the second question of p69 basic training of Mathematics (Shanghai Education Press) in Volume 1 of Grade 7

6 5 -1 1 x=y=15

The resultant force of the two forces F1 and F2 is F. if the angle between the two forces remains unchanged and one of the forces becomes larger, can the magnitude of F remain unchanged?
Please give your reasons or give an example

The size may not change, but the direction will change. For example, F1 = 3N, F2 = 1n, the direction is opposite, the resultant force is 2n, but when F2 becomes 5N, the reasonable value is still 2n, but the direction is opposite