High 1 Math Problem Ask for Help 10

Change the problem and don't understand.

What is this symbol?

Note (1) asks that the Chinese formula is <1>

1> Multiply 3 on both sides to get < 2> formula:

3a1 + 3^2a2 + 3^3a3 + 3^(n-1)an-1 + 3^(n)an=n

Subtract the <2> wrong one place (right shift) with the <1 > alignment to get <4>:

a1 + 3*( a2 - a1 ) 3^2*( a3 - a2 ) 3^(n-1)*(an - an-1 ) 3^n*an = -2/3*n

Note bn = an - an-1, a0 = 0, a1 = 1 3, b1 = a1 = 1 3

4> is formulated as:

b1 + 3*b2 + 3^2*b3 + 3^( n-1)*bn - 3^n*an = -2/3*n

The first n terms are proportional series, the common ratio is 3, and the sum of them is found sn

b1*( 1 - 3^n )/( 1 - 3 ) 3^n*an = -2/3*n

Bring in b1 and simplify it to get an:

3^( 1 - n)*(2n + 3^n/2 - 1/2).over!

The second question is not difficult, figure it out for yourself

Question 1: According to Veda's theorem, x +x = k + 1 and x x = k 4 + 1

s=(x₁)²x₂)²x₁+x₂)²2x₁x₂=(k+1)²-2(k²/4+1)=1/2(k²+4k-2)=1/2[(k+2)²-6]

Since (k+2) 0, the minimum value obtained at k=-2 is 1 2 (-6)=-3

Question 2: You and v are two different real roots of the equation x-1 x=1 (i.e., x -x-1=0).

According to Veda's theorem, u+v=1, uv=-1

So u v+v u=(u +v ) uv=[(u+v) -2uv] uv=(u+v) uv-2=-3

Solution: 3 Even function 2, [k - 3, k + 6] (k n 2

Hope it helps!

Question 1: 2 3 odd functions.

Thank you for giving me a few points.

5.Just follow the steps in the book.

6.Convergence.

Analysis: y1 is proportional to x2, y2 is inversely proportional to x+3, let y1=k1x 2, y2=k2 (x+3).

y=y1+y2=k1x^2+k2/(x+3)

by y=2 when x=0; When x=3, y=0, k2 3=2And 9k1+k2 6=0, k1=-1 9,k2=6,y=-x 2 9+6 (x+3), where x≠3,2, solution: obtained by the principle of leverage, 300*1*sin45°=f*d*sin45°, code eggplant.

f = 300 d, (d 0 m).

When delayed d=, f=300, f(x)=3x 2+2xf'(2),f'(x)=6x+2f'(2),f'(2)=12+2f'(2), i.e. f'(2)=-12.

f(x)=3x^2-24x,f'(x) = 6x-24, hence f'(5)=6.

The key is to solve the differential equations, but also pay attention to f(0) 1