Question

The cubes of the natural numbers are grouped as $1^3,(2^3,3^3),(4^3,5^3,6^3)...$ then sum of the numbers in the nth group is

• A. $\frac{n^2}{12}(n^2+1)(n^2+4)$
• B. $\frac{n^3}{8}(n^2+1)(n^2+3)$
• C. $\frac{n^3}{8}(n^2+1)(n^2+4)$
• D. $\frac{n^2}{16}(n^2+1)(n^2+4)$

Answer 1

Answer: (B)

$\frac{n^3}{8}(n^2+1)(n^2+3)$

Series using the last element of the each group is
$1^3,3^3,6^3,{10}^3,...$
$\therefore t_n$ of $1,3,6,10,...$ is $\frac{1}{2}n(n+1)$
Now
if we replace $n$ by $n-1$ we get the last element of ${(n-1)}^{th}$ group we add 1 in it and cubing the number that
number be the first element of $n^{th}$ group. So the
terms in the $n^{th}$ group are
${(\frac{n(n-1)}{2}+1)}^3,{(\frac{n(n-1)}{2}+2)}^3,...{\left(\frac{n(n+1)}{2}\right)}^3$
$\therefore$ Sum of the element of $n^{th}$ group is
$=$sum of the cubes of first $\frac{n(n+1)}{2}$
natural numbers $-$ sum of the cubes of first $\frac{n(n-1)}{2}$ natural numbers
$=\frac{{\left(\frac{n(n+1)}{2}\right)}^2{(\frac{n(n+1)}{2}+1)}^2}{4}-\frac{{\left(\frac{n(n-1)}{2}\right)}^2{(\frac{n(n-1)}{2}+1)}^2}{4}$
$=\frac{n^3}{8}\left[(n^2+1)(n^3+3)\right]$ (on simplification)