Question

The smallest positive integer n, for which $n! < \left( \frac{n + 1}{2} \right)^{n}$ hold is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Let P(n) : $n! < \left( \frac{n + 1}{2} \right)^{n}$

Step I : For n = 2

$\Rightarrow$ $2! < \left( \frac{2 + 1}{2} \right)^{2}$

$\Rightarrow$ $2 < \frac{9}{4}$

$\Rightarrow$ $2 < 2.25$ which is true. Therefore, P(2) is true.

Step II : Assume that P(k) is true, then p(k) : $k! < \left( \frac{k + 1}{2} \right)^{k}$

Step III : For n = k + 1,

$P(k + 1):(k + 1)! < \left( \frac{k + 2}{2} \right)^{k + 1}$

$\Rightarrow$ $k! < \left( \frac{k + 1}{2} \right)^{k}$

$\Rightarrow (k + 1)k! < \frac{(k + 1)^{k + 1}}{2^{k}}$

$\Rightarrow (k + 1)! < \frac{(k + 1)^{k + 1}}{2^{k}}$ …..(i) and

$\frac{(k + 1)^{k + 1}}{2^{k}} < \left( \frac{k + 2}{2} \right)^{k + 1}$ …..(ii)

$\Rightarrow$ $\left( \frac{k + 2}{k + 1} \right)^{k + 1} > 2$ $\Rightarrow$ $\left\lbrack 1 + \frac{1}{k + 1} \right\rbrack^{k + 1} > 2$

$\Rightarrow 1 + (k + 1)\frac{1}{k + 1} +^{k + 1}C_{2}\left( \frac{1}{k + 1} \right)^{2} + ........ > 2$

$\Rightarrow 1 + 1 +^{k + 1}C_{2}\left( \frac{1}{k + 1} \right)^{2} + ...... > 2$

Which is true, hence (ii) is true.

From (i) and (ii), $(k + 1)! < \frac{(k + 1)^{k + 1}}{2^{k}} < \left( \frac{k + 2}{2} \right)^{k + 1}$

$\Rightarrow$ $(k + 1)! < \left( \frac{k + 2}{2} \right)^{k + 1}$

Hence $P(k + 1)$ is true. Hence by the principle of mathematical induction P(n) is true for all $n \in N$

Trick : By check option

(1) For n = 1, $1! < \left( \frac{1 + 1}{2} \right)^{1}$ ⇒ $1 < 1$ which is wrong

(2) For n = 2, $2! < \left( \frac{3}{2} \right)^{2}$ ⇒ $2 < \frac{9}{4}$ which is correct

(3) For n = 3, $3! < \left( \frac{3 + 1}{2} \right)^{3}$ ⇒ 6 < 8 which is correct

(4) For n = 4, $4! < \left( \frac{4 + 1}{2} \right)^{4}$ ⇒ $24 < \left( \frac{5}{2} \right)^{4}$

⇒ 24 < 39.0625 which is correct.

But smallest positive integer n is 2.