Question

write the set builder form of the following set B={2,5,10,17,26,37,.........,101​

Answer 1

Answer:

{ X:X belongs to natural numbers adding odd number to the preceding term such as 3 ,5,7,9,.......}

Answer 2

First let me make a general form for the given sequence.

Here each consecutive term differs by consecutive odd numbers, i.e., the differences between two consecutive terms form sequences of odd numbers except 1.

5 - 2 = 3

10 - 5 = 5

17 - 10 = 7

26 - 17 = 9

and so on.

But we note other thing that,

5 - 2 = 3

10 - 2 = 3 + 5

17 - 2 = 3 + 5 + 7

26 - 2 = 3 + 5 + 7 + 9

and so on.

In general, for n'th term in the sequence $a_n\ \textgreater\ a_1=2,$

$a_n-2=\sum_{k=1}^{n-1}(2k+1)\\\\\\a_n=2+2\sum_{k=1}^{n-1}k+\sum_{k=1}^{n-1}1\\\\\\a_n=2+2\cdot\dfrac {n(n-1)}{2}+n-1\\\\\\a_n=2+n(n-1)+n-1\\\\\\a_n=2+n^2-n+n-1\\\\\\a_n=n^2+1$

Let's check whether it's true.

1² + 1 = 2

2² + 1 = 5

3² + 1 = 10

4² + 1 = 17

And the last term in the set is,

10² + 1 = 101

Thus each term in the set can be written in the form n² + 1. Here 'n' is any natural number less than or equal to 10.

Hence the set in builder form is,

$\large\underline {\underline {\text { B=\left\{x\ :\ x=n^2+1,\ n\in\mathbb{N},\ n\leq10\right\} }}}$