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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.
 

Posted by

Anam Khan

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