Question

Prove: $\left( 2n+7\right) <\left( n+3\right)^{2} ,\ for\ n\in N$

Answer 1

Hint: In this question it is given that we have to prove that $\left( 2n+7\right) <\left( n+3\right)^{2} $, where $n\in N$. So to prove this statement we have to use the principle of mathematical induction, i.e, we have to show that this statement is true for n=1, if it is true then we have to assume that this statement is also true for n=k, and by using this assumption we have to prove this is also true for n=k+1.
If the above steps satisfy then only we can say that our given statement is also true.

Complete step-by-step answer:
Let us consider the given statement as p(n),
p(n):$\left( 2n+7\right) <\left( n+3\right)^{2} $, for $n\in N$.

For n=1,
L.H.S
$$\left( 2n+7\right) =2\times 1+7=9$$
R.H.S
$\left( n+3\right)^{2} =\left( 1+3\right)^{2} =4^{2}=16$
Since 9<16, $\Rightarrow$ L.H.S < R.H.S
Therefore, p(n) is true for n=1.
Now let us assume that p(n) true for n=k,
p(k):$\left( 2k+7\right) <\left(k+3\right)^{2} $,..........(1)
We have to prove that p(n) is true for n=k+1,
 i.e, p(k+1):$\{ 2\left( k+1\right) +7\} <\{ \left( k+1\right) +3\}$ is true.
Now from statement (1), we have,
$\left( 2k+7\right) <\left(k+3\right)^{2} $
$\Rightarrow \left( 2k+7\right) +2<\left(k+3\right)^{2} +2$ [adding 2 on the both side]
$\Rightarrow \left( 2k+2+7\right) <\left( k+3\right)^{2} +2$
$\Rightarrow \{ 2\left( k+1\right) +7\} < k^{2}+2\times 3\times k+3^{2}+2$
$\Rightarrow \{ 2\left( k+1\right) +7\} < k^{2}+6k+11$
Now by adding (2k+5) in the both side, we get,
$\{ 2\left( k+1\right) +7\} +\left( 2k+5\right) < k^{2}+6k+11+\left( 2k+5\right) $
$\Rightarrow \{ 2\left( k+1\right) +7\} +\left( 2k+5\right) < k^{2}+8k+16$
$\Rightarrow \{ 2\left( k+1\right) +7\} +\left( 2k+5\right) < k^{2}+2\times 4\times k+4^{2}$
$\Rightarrow \{ 2\left( k+1\right) +7\} +\left( 2k+5\right) <\left( k+4\right)^{2} $
$\Rightarrow \{ 2\left( k+1\right) +7\} +\left( 2k+5\right) <\{ \left( k+1\right) +3\}$
So from the above we can say that if $$\{ 2\left( k+1\right) +7\} +\left( 2k+5\right) <\{ \left( k+1\right) +3\}$$ then, $$\{ 2\left( k+1\right) +7\} <\{ \left( k+1\right) +3\}$$.
Therefore, p(k+1) is true.
So our given statement is correct, i.e, $\left( 2n+7\right) <\left( n+3\right)^{2} $, where $n\in N$,
Hence proved.


Note: To verify this type of inequation you must know the ‘Principle of Mathematical Induction’. And Mathematical induction states that if the statement is true for n=1 then it must be true for n=k and after that by using this we have to prove that this statement is also true for n=k+1.
So by the above process one can conclude that the given statement is true.