Show that (1 + x)−1 = 1 + x + x2 + ... in the power series ring R[[x]].?

Answer 1

Multiply 1 + x + x^2 + ... by (1+x).
1 + x + x^2 + ... = Σ(n=0 to ∞) 1.x^n
So (1 + x + x^2 + ... ) (1-x)
= [Σ(n=0 to ∞) 1.x^n] (1-x)
= Σ(n=0 to ∞) (1.x^n)(1-x)
= Σ(n=0 to ∞) 1.x^n - 1.x^(n+1)
= Σ(n=0 to ∞) 1.x^n - Σ(n=0 to ∞) 1.x^(n +1)
= Σ(n=0 to ∞) 1.x^n - Σ(n=1 to ∞) 1.x^n
= 1.x^0 + Σ(n=1 to ∞) 1.x^n - Σ(n=1 to ∞) 1.x^n
= 1 + Σ(n=0 to ∞) (1.x^n - 1.x^n)
= 1 + 0
= 1.

Answer 2

BY taylor theorem if f(x) is deffrentiable infinite times then
f(x)=sigma f^(0)/n![x^n]. If we put g(x)=(1 + x)−1 then [g^(n)(x)]=((-1)^(n+1)/n!)[(1 + x)^((n+1)], and it should be mentioned that U made a mistake It is the serie of (1-X^(-1))

Answer 3

(x + a million)² = 9 (x + a million)² - 9 = 0 (x + a million)² - 3² = 0 ? you note of: a² - b² = (a + b)(a - b) (x + a million + 3)(x + a million - 3) = 0 (x + 4)(x - 2) = 0 First case: (x + 4) = 0 ? x = - 4 2d case: (x - 2) = 0 ? x = 2 answer = { - 4 ; 2 } ? answer B