Prove By Induction The Inequality 1+xn≥ 1+n X, Whenever X Is ...

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Let p(n) be the statement given by

P(n):(2+x)n≥1+nx

For n=1,P(1):(1+x)1≥1+x

∵1+x=1+x

∴P(1)is true.

Assume that , P(k) is true for some natural number k.

Then P(k):(1+x)k≥1+kx

We shall now show that . P(k+1)is true, whenever P(k) is true.

i.e. P(k+1)k+1≥1+(k+1)x

Now consider P(K) is true.

∴(1+x)k≥1+kx

⇒(2+x)k(1+x)≥(1+k)(1+x)

[multiplying both sides by (1+x)]

⇒(1+x)k+1≥1+(k+1)x+kx2

⇒(1+x)k+1≥1+(k+1)(x+kx2≥1+(k+1)x

[∵kx2>0]

⇒(1+x)k+1≥1+(k+1)x

⇒P(k+1)is true.

Hence, by the principle of mathematical induction P(n) is true,∀n∈N.

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