Question

Is there a way to form an equation of this sequence?? `1,3,6,10,15,21...` The equation for the `n`th term

Answer 1

`T_n = 1/2n(n+1)`

Explanation:

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 = color(white)((0)+)1+color(white)((0)+)2+...+(n-1)+color(white)((0)+)n`

`color(white)(2T_n) +color(white)((0)+)n+(n-1)+...+color(white)((0)+)2+color(white)((0)+)1`

`color(white)(2T_n) = (n+1)+(n+1)+...+(n+1)+(n+1)`

`color(white)(2T_n) = n(n+1)`

So:

`T_n = 1/2 n(n+1)`

Why are they called triangular numbers?

`1color(white)(ooooo)3color(white)(ooooo)6color(white)(oooooo)10color(white)(0/0)...`
`ocolor(white)(ooooo)ocolor(white)(ooooo)ocolor(white)(ooooooo)o`
`color(white)(ooooo)ocolor(white)(o)ocolor(white)(ooo)ocolor(white)(o)ocolor(white)(ooooo)ocolor(white)(o)o`
`color(white)(oooooooooo)ocolor(white)(o)ocolor(white)(o)ocolor(white)(ooo)ocolor(white)(o)ocolor(white)(o)o`
`color(white)(ooooooooooooooooo)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`

Method of differences

The method of differences is a more general method for finding the formula of the general term of a polynomial sequence...

Write down the given sequence:

`color(red)(1), 3, 6, 10, 15, 21`

Write down the sequence of differences between successive terms:

`color(green)(2), 3, 4, 5, 6`

Write down the sequence of differences of those differences:

`color(blue)(1), 1, 1, 1`

Having reached a constant sequence, we can write down a formula for the `n`th term using the first term of each of these sequences as coefficients...

`a_n = color(red)(1)/(0!)+color(green)(2)/(1!)(n-1)+color(blue)(1)/(2!)(n-1)(n-2)`

`color(white)(a_n) = 1+2n-2+1/2n^2-3/2n+1`

`color(white)(a_n) = 1/2n^2+1/2n`

`color(white)(a_n) = 1/2n(n+1)`