Prove the following by using the principle of mathematical induction for all
Question:

Prove the following by using the principle of mathematical induction for all $n \in \mathrm{N}$ : 

$(2 n+7)<(n+3)^{2}$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n):(2 n+7)<(n+3)^{2}$

It can be observed that $P(n)$ is true for $n=1$ since $2.1+7=9<(1+3)^{2}=16$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$(2 k+7)<(k+3)^{2}$ (1)

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

Consider

$\{2(k+1)+7\}=(2 k+7)+2$

$\therefore\{2(k+1)+7\}=(2 k+7)+2<(k+3)^{2}+2$ [using (1)]

$2(k+1)+7<k^{2}+6 k+9+2$

 

$2(k+1)+7<k^{2}+6 k+11$

Now, $k^{2}+6 k+11<k^{2}+8 k+16$

$\therefore 2(k+1)+7<(k+4)^{2}$

 

$2(k+1)+7<\{(k+1)+3\}^{2}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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