Question

Assertion :STATEMENT-1 : $1.3.5.....(2n-1)>n^n,nϵN$ Reason: STATEMENT-2 : The sum of the first $n$ odd natural numbers is equal to $n^2$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is D Statement-1 is False, Statement-2 is True

$1+3+5+...+(2n-1)=\frac{n}{2}(2+(n-1\left)2\right)=n^2$
$\therefore$ Statement 2 is true.
Now, applying $AM\ge GM$ to the n odd numbers
$\frac{1+3+...+(2n-1)}{n}\ge \left(\mathrm{1.3.5...}\right(2n-1){)}^{\frac{1}{n}}$
$\Rightarrow \frac{n^2}{n}\ge \left(\mathrm{1.3.5...}\right(2n-1){)}^{\frac{1}{n}}$
or $\left(\mathrm{1.3.5...}\right(2n-1\left)\right)\le n^n$
Thus Statement 1 is False