If x ^ 3 + x ^ 2 + X + 1 = 0, find the value of algebraic formula 1 + X + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 +... + x ^ 2006 + x ^ 2007

(2007+1)÷4=502∴1+x+x^2+x^3+x^4+x^5+...+x^2007+x^2008-x^2008=1+x+x²+x³+x^4(1+x+x²+x³)+x^8(1+x+x²+x³)+.+x^2004(1+x+x²+x³)=0+0+...0=0

When x is taken as 1 / 2007,1 / 2006,... 1 / 2,1,2,3,... 20062007 respectively, find out the square of the algebraic formula 1-x / the square of 1 + X, and find the sum

Excuse me, this is a math problem in senior high school? If you are a sophomore in senior high school, you should be able to guess. Here's a hint: the value of X is 1 / 2007 + the value of X is 2007 ∑ is 1. The formula is: x square / (1 + x square) + 1 / (1 + x square). Do the rest by yourself

22 + 0.44 + 0.66 + 0.88 + 1.1 + 1.12 + 2.24 + 3.36 + 4.48 + 5.6

Original form
=0.22*(1+2+3+4+5)+1.12*(1+2+3+4+5)
=(0.22+1.12)*(1+2+3+4+5)
=1.34*15
=20.1

If n is any real number, the value of (n + 11) ^ 2-N ^ 2-N ^ 2 can always be divided by K, then K is equal to?

You have made a mistake in the question. Check it yourself

1.8-8x/1.2-1.3-3x/2-5x-0.4/0.3=0

x=(1/29)

(14.2 + 14.1 + 14.4 + 14.3 + 14.5) * how to calculate 0.2 with a simple method

14.3*5*0.2=14.3*1=14.3
Note: 14.3 is the average of (14.2 + 14.1 + 14.4 + 14.3 + 14.5)

Let the sum of the first n terms of the sequence an be Sn, A1 = a, an + 1 = Sn + 3 ^ n, if an + 1 > = an, the value range of a is obtained

an+1=Sn+3^n;
an=Sn-1+3^(n-1);
By subtracting the two formulas, we can get an + 1 = 2An + 2 * 3 ^ (n-1) > = an
So an > = - 2 * 3 ^ (n-1)
a2=2(a1+1)=2a1+2=2a+2>=a a>=-2
n> 2, a > = - 2 satisfies an > = - 2 * 3 ^ (n-1), so the value range of a is [- 2, + ∞)

If x ^ 3 + x ^ 2 + X + 1 = 0, find the value of algebraic formula 1 + X + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 +... + x ^ 2006 + x ^ 2007