( 1)f(x)=x？ +n？ x- 1 .

Note that f'(x)=3x? +n？ > 0, so f(x) is written on R. ..

And f (0) =- 1 < 0, f( 1)=n? >0.

Therefore, f(x) has a unique zero an at (0, 1).

∴an∈(0, 1)。

(2)

f(an)= 0 & lt； => Ann? +n？ an- 1=0。

& lt=> An (An? +n？ )= 1

& lt= & gtan= 1/(an？ +n？ )

Ann again? +n？ < 1+n？ < n+n？ =n(n+ 1).

Therefore, an >1/n (n+1) =1/n-1/(n+1).

Then a1+a2+...+an > ∑ (1/n-1(n+1)) =1-1(n+/kloc-)

Where's Ann? +n？ > n？ , so an < 1/n? .

And then a1+a2+...+an < ∑1/n? .

From the geometric meaning of definite integral, ∑ 1/n? < 1+(2，n)∫( 1/n？ )dn = 1+(- 1/n+ 1/2)= 3/2- 1/n < 3/2。

2 1.(②)

( 1)f(x)=√( 1+x？ ), it is obvious that f(x) decreases on R- and increases on R+.

Its image is hyperbolic y? -x？ = 1, f(x) is an even function.

L ≥| f (x 1)-(x2) |/(| x1-x2 |) holds for any x1,x2 ∈ r.

According to symmetry, l is greater than or equal to the tangent slope of any point on the image.

The asymptote of this hyperbola is y = x and the maximum slope is 1, so L≥ 1.

(2)

①

Note that a(n+ 1)=f(an), and a(n+2)=f(a(n+ 1)).

Let x 1=an and x2=a(n+ 1).

Then | f (a (n+1))-f (an) | =| a (n+2)-a (n+1) |≤ l | a (n+1)-an |.

If bn=|a(n+ 1)-an|, then b(n+ 1)≤Lbn.

So bn≤Lb(n- 1)≤L? b(n-2)≤...≤L^(n- 1)×b 1.

Then ∑ bn ≤∑ l (n-1) × b1= b1(1-l n)/(1-l) ≤ b1(/kloc)

∴ ① Established.

I haven't thought about this for the time being. ..

20.

( 1)f(0)=0=c .

f(- 1/2+x)= f(- 1/2-x)& lt； = & gtX=- 1/2 is the symmetry axis of f(x).

And f(x)=ax? +ax≥x is a constant, then ax? +(a- 1)x≥0 holds.

So △≤0, then (a- 1)? ≤0。 Get a= 1.

So f(x)=x? +x .

(2)g(x)=x？ +x-|λx- 1| .

(1) when x ≥ 1/λ.

g(x)=x？ +( 1-λ)x+ 1。

When λ≥2, g(x) decreases at [1/λ, (λ- 1)/2] and increases at (λ- 1)/2, +∞).

When 0 < λ < 2, g(x) increases to (1/λ, +∞).

② When x < 1/λ.

g(x)=x？ +(λ+ 1)x- 1。

At this time, g(x) decreases at (-∞, -(λ+ 1)/2) and increases at (-(λ+ 1)/2, 1/λ).

(3)

① 0 < λ≤ 1.

According to (2), when x∈(0, 1), g(x) increases at (0, 1), and g(x)=x? +(λ+ 1)x- 1 .

And g (0) =- 1 < 0, g (1) = λ+ 1 > 0. Therefore, g(x) has a unique zero at (0, 1).

② λ > 1.

(I) when x ∈ (0, 1/λ), g(x)=x? +(λ+ 1)x- 1, and g(x) increases by (0, 1/λ).

G(0)=- 1 0. So g(x) has a unique zero at (0, 1/λ).

(ii) when x ∈ (1/λ, 1), g(x)=x? +( 1-λ)x+ 1 .

When {1} 1 < λ < 2, g(x) increases at (1/λ, 1).

G (1/λ) > 0, so at this time g(x) has no zero at (1/λ, 1).

When {2} 3 > λ≥ 2, g(x) decreases at [1/λ, (λ- 1)/2] and increases at ((λ- 1)/2, 1).

And g ((λ-1)/2) =1-((λ-1)/2)? > 0, so g(x) has no zero at this time.

{3} When λ≥ 3, g(x) decreases at (1/λ, 1), while g( 1)=3-λ≤0.

So at this time, g(x) has a unique zero at (1/λ, 1).

To sum up, when 0 < λ < 3, g(x) has zero at (0, 1); When λ≥3, g(x) has two zeros at (0, 1).