Question

The sequence of natural numbers is divided into classes as follows
Prove that the sum of the numbers in the nth row is $n(2n^2+1)$

Answer 1

The number of numbers in successive rows are 2, 4, 6 and hence there will be 2n numbers in $n^{th}$ row which will be in A.P. of common difference 1. The first term in successive rows are
S = 1 + 3 + 7 + 11......+ $t_n$
S = 1 + 3 + 7 + 11......+ $t_{n-1}-t_n$
Subtracting,
$\therefore$ 0 = 1 + [2 + 4 + 6 + (n - 1) terms] - $t_n$
$\therefore t_n=1+\frac{2(n-1)n}{2}=n^2-n+1$
$\therefore S_n$ = sum of an A.P in which
A = $n^2$ - n + 1, D = 1, N = 2n.
$\therefore S_n=\frac{2\pi }{2}\left[2\right(n^2-n+1\left)\right(2n-1)\cdot 1]$
$=n(2n^2+1)$