Question

The series of natural numbers is divided into groups $\left(1\right);(2,3,4)$; $(5,6,7,8,9)$; __________ and so on. Show that the sum of the numbers in the nth groups is $(n-1{)}^3+n^3$.

Answer 1

The number of terms in successive groups are $1,3,5$,______ and hence in nth group will be nth term of this A.P. i.e., $2n-1=N$
The last term of successive groups are $1^2,2^2,3^2$,... and hence of nth group is $n^2$ and of $(n-1)$th group is $(n-1{)}^2$. Hence $1$st term of nth group is one more than the last term of $(n-1)$th group
$\therefore A=(n-1{)}^2+1=n^2-2n+2$
Also terms in each group are in A.P. whose D$=1$.
$\therefore$ Sum of terms in nth group is sum of an A.P.
$=\frac{N}{2}[2A+(N-1\left)D\right]$
$=\frac{2n-1}{2}[2n^2-4n+4+(2n-2)\cdot 1]$
$=(2n-1)(n^2-n+1)$
$=2n^3-3n^2+3n-1=n^3+(n-1{)}^3$.