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

Option: 2
2

Option: 3
3

Option: 4
4

Answers (1)

best_answer

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

Step I : For n = 2 $\Rightarrow \quad 2 !<\left(\frac{2+1}{2}\right)^2 \Rightarrow 2<\frac{9}{4}$

$\Rightarrow \quad 2<2.25$ 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, $P(k+1):(k+1) !<\left(\frac{k+2}{2}\right)^{k+1}$

$\begin{aligned} & \Rightarrow \quad k !<\left(\frac{k+1}{2}\right)^k \Rightarrow(k+1) k !<\frac{(k+1)^{k+1}}{2^k} \\ & \Rightarrow \quad(k+1) !<\frac{(k+1)^{k+1}}{2^k} \end{aligned}$ ......(i)

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

$\begin{aligned} & \Rightarrow\left(\frac{k+2}{k+1}\right)^{k+1}>2 \Rightarrow\left[1+\frac{1}{k+1}\right]^{k+1}>2 \\ & \Rightarrow 1+(k+1) \frac{1}{k+1}+{ }^{k+1} C_2\left(\frac{1}{k+1}\right)^2+\ldots \ldots . .>2 \\ & \Rightarrow 1+1+{ }^{k+1} C_2\left(\frac{1}{k+1}\right)^2+\ldots \ldots>2 \end{aligned}$

Which is true, hence (ii) is true.

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

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

Hence is true. Hence by the principle of mathematical induction P(n) is true for all

Trick : By check options,

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

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

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

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

$\Rightarrow 24<39.0625$ which is correct.

But smallest positive integer n is 2.

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Anam Khan

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