Question

The natural numbers are written in the form of a triangle as shown below :
Find the sum of the numbers in the nth row

Answer 1

$n^{th}$ row will have n elements which will form an A.P. of common difference 1, we have to determine only it's 1st term.
Successive first terms are 1, 2, 4, 7, 11 ,.....
1st difference are 1, 2, 3, 4...
2nd differences 1, 1, 1, .....equal.Hence $n^{th}$ term will be of the form
$T_n=a{n}^2+bn+c$
$T_1=a+b+c=1$
$T_2=4a+2b+c=2$
$T_3=9a+3b+c=4$
$\therefore =3a+b=1,5a+b=2$
$a=\frac{1}{2},b=-\frac{1}{2},c=1$
$T_n=\frac{1}{2}(n^2-n+2)$
$\therefore Nowa=\frac{1}{2}(n^2-n+2),d=1,n=n$
$\therefore S=\frac{n}{2}\left[\right(2a+(n-1)d]$
$\therefore S=\frac{n}{2}[n^2-n+2+n-1]=\frac{n}{2}(n^2+1)$