The arithmetic mean of 1, 8, 27, 64, … up to n terms is given by

Concept:

Sum of cubes of first n natural numbers:\(\sum {{\rm{n}}^3} = {1^3} + {2^3} + {3^3} + \ldots + {{\rm{n}}^3} = {\rm{\;}}{\left[ {\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}} \right]^2}\)

The arithmetic mean is the sum of all the numbers in a data set divided by the quantity of numbers in that set.

Calculation:

We have to find the arithmetic mean of 1, 8, 27, 64, … up to n terms,

{1, 8, 27, 64, … up to n terms} = {13, 23, 33, 43 … up to n terms}

Now,

Arithmetic mean = AM = \(\frac{{{1^3} + {2^3} + {3^3} + \ldots + {{\rm{n}}^3}}}{{\rm{n}}}\)
\(\Rightarrow {\rm{AM}} = {\rm{\;}}\frac{{{{\left[ {\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}} \right]}^2}{\rm{\;}}}}{{\rm{n}}} = \frac{\rm n^2\;(n+1)^2}{4\rm n}\)

\(\Rightarrow {\rm{AM}} = {\rm{\;}}\frac{{{\rm{n}}{{\left( {{\rm{n}} + 1} \right)}^2}}}{4}\)

∴ Option 3rd is correct answer.