Question

what is the set builder from for A={1,3,6,10}
who will say first correct answer I will it as brainlist answer​

Answer 1

Consider the elements of the set A.

1, 3, 6, 10

These are the first 4 triangular numbers!

The general formula for finding triangular numbers is $\dfrac{n(n+1)}{2},\ \ \forall n\in\mathbb{N}.$

Here,

$1=\dfrac{1\cdot 2}{2}\ \ \ ; \ \ \ 3=\dfrac{2\cdot 3}{2}\ \ \ ; \ \ \ 6=\dfrac{3\cdot 4}{2}\ \ \ ; \ \ \ 10=\dfrac{4\cdot 5}{2}$

Here the least value of 'n' is 1 and the highest value of 'n' is 4.

Hence A in set builder form can be,

$\large\text{ A=\left\{x:x=\dfrac{n(n+1)}{2},\ n\in\mathbb{N},\ 1\leq n\leq 4\right\} }$

OR

In terms of combination, we can write the general form of triangular numbers as,

$\dfrac{n(n+1)}{2}=\dfrac{(n-1)!\ n(n+1)}{2(n-1)!}=\dfrac{(n+1)!}{2(n-1)!}=\dfrac{(n+1)!}{2!(n+1-2)!}=\ ^{n+1}\!C_2$

We can apply this. Here the least and the highest values of 'n' are also 1 and 4 respectively.

Hence A in set builder form can be,

$\large\text{ A=\left\{x:x=\ ^{n+1}\!C_2,\ n\in\mathbb{N},\ 1\leq n\leq 4\right\} }$

OR

Since the four elements of the set A are triangular numbers, we can write,

1 = 1

3 = 1 + 2

6 = 1 + 2 + 3

10 = 1 + 2 + 3 + 4

That is, each triangular number can be written as the sum of first consecutive natural numbers. So, we can find out that,

$\displaystyle \frac{n(n+1)}{2}=\sum_{k=1}^nk$

Hence A in set builder form can be,

$\large\text{ A=\left\{x:x=\displaystyle\sum_{k=1}^nk,\ n\in\mathbb{N},\ 1\leq n\leq 4\right\} }$