How do you calculate the mathematical formula f x 1 x 0 2 x 1 n x n 1?

The above is the high school method, I forgot what it is called, the principle is to multiply by the same number, so that the formula moves to the right, but how are you infinite, is n = infinite? So let's use the second method, which is the university method, using progression, and I'm not done yet.

f(x)=1*x^0+2*x^1+..n*x^(n-1)+.

xf(x)= x +.n-1)x^(n-1)+nx^n+..

f(x)-xf(x)=1+x+x^2+x^3+..x^(n-1)-nx^n+..

1-x)f(x)=(1-x^n)/(1-x)-nx^n+..

f(x)=(1-x^n)/(1-x)^2 - nx^n/(1-x)+.x is not equal to 1

When 01, the root of the pole

When x=1

f(x)=1+2+..n-1)+n...=n(n+1)/2+..

There should be a lot of after-class exercises for questions like this, but it shouldn't be.

As a hint, if f(x)*x, what happens to the right?

Look to the right of f(x) again and see if there is any inspiration.

ps: If you come up with it yourself, it will be your own; If you want to learn math well, you must be diligent in thinking.

b: Jump break.

It is continuous in each segment interval, but there are discontinuities between intervals.

The area of the f(x) image is divided into i small rectangles from the x-axis, and the width of each image is 1 n.

The height of the ith part is f(i n).

The area of the ith part is then 1 n times f(i n).

After summing, there is a definite integral.

The definite integral is the area between the f(x) image and the x-axis.

This question f(x)=1 (1+x).

It is a summation symbol, English name: sigma, Chinese name: Sigma (uppercase, lowercase).

The eighteenth Greek letter. In Greek, if the last letter of a one-word jujube pants is lowercase sigma, write the letter as , which is also called final sigma (unicode: u+03c2).

In modern Greek numerals represent 6.

In mathematics, we use as a summation symbol; Use lowercase letters to indicate the standard Xunyan bucket difference.

Applications in Physics:

In physics, we use its lowercase letters to denote areal density. (Correspondingly, the volume density is denoted and the line density is denoted by the linear density). Areal density is the mass per unit area of a substance of specified thickness in terms of engineering materials.

In chemistry, we use its lowercase letter to denote a type of covalent bond. A covalent bond formed by the overlap of two atomic orbitals along the axis of orbital symmetry leads to an increase in the probability of electrons appearing between nuclei, which is called a bond. The key belongs to the localized bond, which can be either a general covalent bond or a coordination covalent bond.

A single bond in general is a key.

According to the derivative rule of multiplication, there will be a trembling mountain (n+1) formula to add, the first formula does not contain x terms, and the subsequent formulas contain x terms, and they are all zero at zero.

Therefore, the final result of repentance is the order of n.

There is a lack of prosperity in the method of derivation.

Let f(x) = x (2n-1) (2n-1) then f'(x)=∑x^(2n-2)

The common ratio is x dang|x|At <1, there is x (2n-2)=1 (1-x ) which is f'(x)=1/(1-x²)=1/2[1/(1-x)+1/(1+x)]

The integral yields: f(x)=1 2ln[(1+x) (1-x)]+c by the original formula, f(0)=0, so there is f(0)=0+c=0, get: caution c=0 so there is f(x)=1 2ln[(1+x) (1-x)]<

Knowledge Expansion: The sum function is the sum of the infinite series of function terms.

The sum function is the sum of power series. That is, n starts from 1 and goes to positive infinity.

f(x)=y=(x+1)(2x+1)(3x+1).(nx+1), find f'(0)

The natural logarithm of both sides is obtained: lny=ln(x+1)+ln(2x+1)+ln(3x+1)+.ln(nx+1)

Take the derivative of x on both sides to get y'Change the stupidity with y=1 (x+1)+2 (2x+1)+3 (3x+1)+.n/(nx+1)

Hence f'Nuclear reed(x)=f(x)[1 gear beam (x+1)+2 (2x+1)+3 (3x+1)+.n/(nx+1)]

f'(0)=f(0)(1+2+3+.+n)=f(0)(1+n)n/2

Because f(0)=1, f'(0)=n(n+1)/2.,6,f(x)=(x+1)(2x+1)(3x+1).(nx+1), find f'(0)

f(x)=(x+1)(2x+1)(3x+1)..(nx+1), find f'(0) Thank you very much.

f(x)=(x-1)(x-2)..x-n)(x+1)^(1)(x+2)^(1)..x+n)^(1)

So f'(x)=(x-1)’*x-2)..x-n)(x+1)^(1)(x+2)^(1)+(x-1)(x-2)‘.x-n)(x+1)^(1)(x+2)^(1)..

x+n)^(1)..x+n)^(1)+…x-1)(x-2)..x-n)[(x+1)^(1)]'(x+2)^(1)..

x+n)^(1)+…x-1)(x-2)..x-n)(x+1)^(1)(x+2)^(1)..x+n)^(1)]'

x-2)..x-n)(x+1)^(1)(x+2)^(1)+(x-1)(x-2)‘.x-n)(x+1)^(1)(x+2)^(1)..

x+n)^(1)..x+n)^(1)+…x-1)(x-2)..x-n)[(x+1)^(1)]'(x+2)^(1)..

x+n)^(1)+…x-1)(x-2)..x-n)(x+1)^(1)(x+2)^(1)..x+n)^(1)]'

With the exception of the first term, every other term has x-1, so x=1 is equal to 0

So f'(1)=(1-2)..1-n)=(-1)^(n-1)*(n-1)!

Note a=(x-1)(x-2).x-n）b=(x+1)(x+2)..x+n)

f(x)'=(a'b+ab')/b²

For the derivative of a, if the derivative is left after (x-1), then f'The term (x-1) in (1) is 0

When x=1

f（x）'|x=1= a‘/b= （-1）（-2）……1-n）/(n+1)!

-1) to the power of (n-1) n(n+1).