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To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N

To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N. Note: x>-1. Hence (1+x)>0. Next. To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n N. Prove: (1 + x) n ≥ 1 + nx for n=1. For n = 1 (1 + x) 1 = 1 + x True for n = 1. Next.

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To prove by induction that (1 + x) n ≥ 1 + nx for x > -1, n  N

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  1. To prove by induction that (1 + x)n ≥ 1 + nx for x > -1, n  N Note: x>-1. Hence (1+x)>0. Next (c) Project Maths Development Team 2011

  2. To prove by induction that (1 + x)n ≥ 1 + nx for x > -1, n N Prove: (1 + x)n ≥ 1 + nx for n=1 For n = 1 (1 + x)1 = 1 + x True for n = 1 Next (c) Project Maths Development Team 2011

  3. To prove by induction that (1 + x)n ≥ 1 + nx for x > -1, n  N • Assume true for n = k. • Therefore (1 + x)k ≥ 1 + kx Prove true for n = k + 1 Multiply each side by 1 + x (1 + x)(1 + x)k ≥ (1 + x)(1 + kx) (1 + x)k+1 ≥ 1 + kx + x + kx2 Since (1 + x)k+1 ≥ 1 + kx + x + kx2 (k > 0 then kx2 ≥0 for all x) Therefore (1 + x)k+1 ≥ 1 + kx + x (1 + x)k+1 ≥ 1 + (k+1)x If true for n = k this implies it is true for n = k + 1. It is true n = 1. Hence by induction (1 + x)n ≥ 1 + nx x > -1, n  N. (c) Project Maths Development Team 2011

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