Question

please if you know then answered me otherwise not.​

Answer 1

→ Given,

The number is 19 which is odd.

→ We have to express $19^2$ as the sum of two consecutive positive number.

Solution :

$\pink{ \boxed{\boxed{\begin{array}{cc}\sf \to \: \: we \: know \: that \: : \\ \\ \sf \: the \: sum \: of \: two \: consecutive \:positive \\ \\ \sf \: integers \: is \: \: {n}^{2} = \frac{ {n}^{2} - 1}{2} + \frac{ {n}^{2} + 1}{2} \\ \\ \sf \: \: when \: \: n \: \: is \: odd \: number\end{array}}}}</p><p>$

Now,

$\sf \: {n}^{2} = {(19)}^{2} = 361$

Again,

$\sf \: {19}^{2} = \frac{ {19}^{2} - 1}{2} + \frac{ {19}^{2} + 1}{2} \\ \\ = > {19}^{2} = \frac{361 - 1}{2} + \frac{361 + 1}{2} \\ \\ = > {19}^{2} = 180 + 181$

So,

${19}^{2} = 180 + 181$