Question

Show that the sum of the $m^{th}$ powers of the first $n$ even numbers is greater than $n{(n+1)}^m$, if $m>1$.

Answer 1

According to power mean inequality $QM\ge AM\ge GM\ge HM$

Thus quadratic mean is greater than arithematic mean ,

So , $m^{th}$ power of first $n$ even number are ,

$\frac{{(2^m+4^m+6^m+...+{\left(2n\right)}^m)}^{\frac{1}{m}}}{n}\ge \frac{2+4+6+...+2n}{n}$

$\frac{(2^m+4^m+6^m+...+{\left(2n\right)}^m)}{n}\ge {\left(\frac{2+4+6+...+2n}{n}\right)}^m$

$\frac{(2^m+4^m+6^m+...+{\left(2n\right)}^m)}{n}\ge {\left(\frac{2(1+2+3+...+n)}{n}\right)}^m$

Sum of $n$ natural number is $\frac{n(n+1)}{2}$

$\frac{(2^m+4^m+6^m+...+{\left(2n\right)}^m)}{n}\ge {\left(\frac{2n(n+1)}{2n}\right)}^m$

$(2^m+4^m+6^m+...+{\left(2n\right)}^m)\ge n{(n+1)}^m$

This is true only when $m>1$