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Base case: $n = 1, 2^1 = 2 > 1^2$

Induction hypothesis: $2^n > n^2$

To prove: $2^{n+1} > {(n+1)}^2$

$\text{LHS} = 2. 2^n > 2. n^2 $ (Induction hypothesis)

$\text{RHS} = {(n+1)}^2 = n^2 + 2n + 1 < \text{LHS}$, hence proved.
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Putting the values of n = 1,2,3,4,5,6,7,8 we find

2^>1^2

2^2=2^2

2^3 <3^2

2^4= 4^2

 

2^ 5> 5^2

2^6 >6 ^2

2^7 > 7 ^2

and so on

 

Hence we conclude that the given statement holds good for values n> 5 ( by induction)

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