Question

Is there a way to form an equation of this sequence?? $1,3,6,10,15,\mathrm{21...}$ The equation for the $nth$ term.

Answer 1

These are the triangular numbers - each term in the sequence being the sum of the first $n$ positive integers:
$T_1=1=1$
$T_2=3=1+2$
$T_3=6=1+2+3$
etc.
Notice that:
$2T_n=1+2+...+(n-1)+n+n+(n-1)+...+2+1$
$=(n+1)+(n+1)+...+(n+1)+(n+1)$
$=n(n+1)$
So:
$T_n=\frac{1}{2}n(n+1)$
Why are they called triangular numbers?
Method of differences
The method of differences is a more general method of finding the formula of the general term of a polynomial sequence ...
Write down the given sequence:
$1,3,6,10,15,21$
Write down the sequence of difference between successive terms:
$2,3,4,5,6$
Write down the sequence of differences of those differences:
$1,1,1,1$
Having reached a constant sequence, we can write down a formula for the nth term using the first term of each of these sequences as coefficients...
$a_n=\frac{1}{0!}+\frac{2}{1!}(n-1)+\frac{1}{2!}(n-1)(n-2)$
$=1+2n-2+\frac{1}{2}{n}^2-\frac{3}{2}n+1$
$=\frac{1}{2}{n}^2+\frac{1}{2}n$
$=\frac{1}{2}n(n+1)$.