Question

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

Solution

## According to power mean inequality $$QM \geq AM \geq GM \geq HM$$Thus quadratic mean is greater than arithematic mean ,So , $$m^{th}$$ power of first $$n$$ even number are ,$$\dfrac{\left(2^{m}+4^{m}+6^{m}+...+\left(2n\right)^{m}\right)^{\dfrac{1}{m}}}{n} \geq \dfrac{2+4+6+...+2n}{n}$$$$\dfrac{\left(2^{m}+4^{m}+6^{m}+...+\left(2n\right)^{m}\right)}{n} \geq\left (\dfrac{2+4+6+...+2n}{n}\right)^{m}$$$$\dfrac{\left(2^{m}+4^{m}+6^{m}+...+\left(2n\right)^{m}\right)}{n} \geq \left (\dfrac{2\left(1+2+3+...+n\right)}{n}\right)^{m}$$Sum of $$n$$ natural number is $$\dfrac{n\left(n+1\right)}{2}$$$$\dfrac{\left(2^{m}+4^{m}+6^{m}+...+\left(2n\right)^{m}\right)}{n} \geq \left( \dfrac{2n\left(n+1\right)}{2n}\right)^{m}$$$$\left(2^{m}+4^{m}+6^{m}+...+\left(2n\right)^{m}\right) \geq n\left(n+1\right)^{m}$$This is true only when $$m>1$$  Mathematics

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