# Show by induction that : (1+x)^n ge 1+nx , for x > -1

Lisa Hardin 2022-07-25 Answered
Show by induction that :
$\left(1+x{\right)}^{n}\ge 1+nx$, for x> -1
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Ali Harper
This is very simple just plug -1 in for x

We have step-by-step solutions for your answer!

Let P(n) be the statement that $\left(1+x{\right)}^{n}1+nx$
Basis step :
P(0) is true because $\left(1+x{\right)}^{0}=1$ (=1+0x = 1 )
Inductive step:
Assume that P(k) is true.
i.e., $\left(1+x{\right)}^{k}1+kx$
To prove that P(k + 1) istrue.
i.e., to show that $\left(1+x{\right)}^{k+1}1+\left(k+1\right)x$
Now $\left(1+x{\right)}^{k+1}=\left(1+x{\right)}^{k}\left(1+x\right)$
(1 + kx) (1 +x) (for n = k, we have$\left(1+x{\right)}^{k}1+kx$ )
$=1+x+kx+k{x}^{2}$
$=1+\left(k+1\right)x+k{x}^{2}$
1 + (k + 1)x
$\therefore P\left(k+1\right)$ is true.
Hence from the principleof mathematical induction $\left(1+x{\right)}^{n}1+nx$ is true.

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