Question

Show that $n{(n+1)}^3<8(1^3+2^3+3^3+....+n^3)$.

Answer 1

Sum of cubes of first $n$ natural numbers $=1^3+2^3+3^3+.......n^3=\frac{n^2(n+1{)}^2}{4}$

Assume $n(n+1{)}^3\le 8(1^3+2^3+3^3+.......n^3)$

$\Rightarrow n(n+1{)}^3\le 8(\frac{n^2(n+1{)}^2}{4})\le 0$

$\Rightarrow n(n+1{)}^2(n+1-2n)\le 0$

$\Rightarrow n(n+1{)}^2(1-n)\le 0$

Since $n\ge 1,1-n\le 0$

Also, $(n+1{)}^2\ge 0$

So, our assumption is true.

Hence Proved