If $p (n) : n^{2} < 2^{n}$ is true for $n \in N - {1} .$ then the minimum value of $n =$
(use principle of mathematical induction)

Open in App

Solution

Given :
$p (n) : n^{2} < 2^{n}$ is true for $n \in N - {1}$
So, $P (n)$ is true for $n = 2, 3, \dots$
For $n = 2, P (2) : 4 < 4$ (false)
For $n = 3, P (3) : 9 < 8$ (false)
For $n = 4, P (4) : 16 < 16$ (false)
For $n = 5, P (5) : 25 < 32$ (true)
So, minimum value of $n = 5$

$p (n) : n^{2} < 2^{n}; n \geq 5$ is true for $n = k$
For $k = n + 1$
$p (n + 1) : (n + 1)^{2} < 2^{n + 1}; n \geq 5$
$p (n + 1) : n^{2} + 2 n + 1 < 2 \cdot 2^{n}; n \geq 5$
and $2 n + 1 < 2^{n}; n \geq 5$
$∴$ True for $k = n + 1$

Suggest Corrections

Similar questions

n \in N

(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{(2 n + 1)}{n^{2}}) = (n + 1)^{2}

n \in N .

\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}

n \in N

(1 + \frac{1}{1}) (1 + \frac{1}{2}) (1 + \frac{1}{3}) \dots (1 + \frac{1}{n}) = (n + 1)

n \in N .

1^{2} + 3^{2} + 5^{2} + \dots + (2 n - 1)^{2} = \frac{n (2 n - 1) (2 n + 1)}{3}

2 + 4 + 6 + . . . . . + 2 n = n^{2} + n

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program