Question

If sum of the $m^{th}$ powers of the first $n$ odd numbers is $\lambda$, for all $m>1$, then

• A. $\lambda <n^m$
• B. $\lambda >n^m$
• C. $\lambda <n^{m+1}$
• D. $\lambda \ge n^{m+1}$

Answer 1

Answer: (D)

$\lambda \ge n^{m+1}$

$\because m>1$

$\therefore \frac{1^m+3^m+5^m+...+{(2n-1)}^m}{n}\ge {\left(\frac{1+3+5+...+2n-1}{n}\right)}^m$ ....$\left(1\right)$
$1,3,5,..(2n-1)$ are in A.P
$\therefore S_n=\frac{n}{2}(a+l)=\frac{n}{2}(1+2n-1)=n^2$
Substituting this in $\left(1\right)$ we get,
$\frac{1^m+3^m+5^m+...+{(2n-1)}^m}{n}\ge {\left(\frac{n^2}{n}\right)}^m=n^m$
$\therefore 1^m+3^m+5^m+...+{(2n-1)}^m\ge n^{m+1}$
or $\lambda \ge n^{m+1}$ from the above question.