Question

The mean of the cubes of the first $n$ natural numbers is :

• A. $\frac{n(n+1{)}^2}{4}$
• B. $n^2$
• C. $\frac{n(n+1)(n+2)}{8}$
• D. $(n^2+n+1)$

Answer 1

The correct option is A $\frac{n(n+1{)}^2}{4}$

Sum of the cubes of first $n$ natural numbers
$={\left[\frac{n(n+1)}{2}\right]}^2=\frac{n^2(n+1{)}^2}{4}$
$\therefore$ Mean $=\frac{n(n+1{)}^2}{4}$