Question

Let $a_1,a_2...a_n$ be a set of $n$ positive real number (not all equal),
Let $m$be a real number then the arithmetic mean of $m^{th}$ power is counted as
(i):$\frac{a_1^m+a_2^m+a_3^m+...+a_1^m}{n}>{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m$ if the real number $m\in R-[0,1]$
(ii): $\frac{a_1^m+a_2^m+a_3^m+...+a_1^m}{n}<{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m$ if $m\in (0,1)$ and equality holds if $m=0$ or $m=1$.
On the basis of above information answer the following questions,If sum of the $m^{th}$ powers of first $n$ natural numbers $\lambda \forall m>1$ then

• A. $\frac{\lambda }{2^m}>{(n^{1/m}.n+1)}^m$
• B. $2^m.\lambda >{(n^{1/m}.(n+1))}^m$
• C. $\frac{2^m}{\lambda }>{(n+(n+1))}^m$
• D. None of these

Answer 1

The correct option is B $2^m.\lambda >{(n^{1/m}.(n+1))}^m$

For $m>1$

$A.M.$ of $m^{th}$ powers $>m^{th}$ powers of $A.M$

$\therefore \frac{1^m+2^m+...+n^m}{n}>{\left(\frac{1+2+...+n}{n}\right)}^m$

$\therefore \frac{\lambda }{n}>{\left(\frac{n+1}{2}\right)}^m$

$\therefore 2^m.\lambda >{(n^{1/m}.(n+1))}^m$
Ans: B