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18.03.2013

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x ≥ -1 : (1 + x)n ≥ 1 + nx, nZ +.



.

1. .

n = 0 1 ≥ 1.

.

2. .

kN

(1 + x)k ≥ 1 + kx;

, (1 + x)(k + 1) ≥ 1 + (k + 1)x;

:

(1 + x)(k + 1) = (1 + x)k(1 + x) ≥ (1 + kx)(1 + x) ≥ 1 + kx + x + kx2 ≥ 1 + kx + x = 1 + (k + 1)x.

, . .


, :

  • x ≠ -1, n = 0, n = 1;
  • x = -1, n ≠ 0.



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