Express ${{23}^{2}}$ as sum of two consecutive numbers.

Hint: Here we will first assume the two consecutive numbers and then we will find the addition of the two assumed consecutive numbers and then we will equate its addition with the number given in the question.

Complete step-by-step answer:
Let’s find the value of ${{23}^{2}}$ which is equal to 529.
Now,
Let the two consecutive numbers be $y$ and $y\text{ }+\text{ }1$.
Now, equating addition of these two consecutive numbers with 529.
$y\text{ }+\text{ }y\text{ }+\text{ }1\text{ }=\text{ }529$
$\Rightarrow$ $2y\text{ }+\text{ }1\text{ }=\text{ }529$
$\Rightarrow$ $2y\text{ }=\text{ }528$
$\Rightarrow$ $y\text{ }=\text{ }\dfrac{528}{2}$
$\Rightarrow$ $y\text{ }=\text{ }264$
$\Rightarrow$ $ y\text{ }+\text{ }1\text{ }=\text{ }265$
So the two consecutive numbers are 264 and 265.

Note: There is also an alternative method to find the consecutive numbers.
Given number is ${{23}^{2}}$
Now,
Let $n\text{ }=\text{ }23$
We have to express $n^2$ in the form of two consecutive numbers.
Formula to find the 1st consecutive number $=\text{ }\dfrac{{{n}^{2}}\text{ }-\text{ }1}{2}$ ………………….$\left( 1 \right)$
and 2nd consecutive number $=\text{ }\dfrac{{{n}^{2}}\text{ }+\text{ }1}{2}$ …………….$\left( 2 \right)$
Let’s evaluate equation $\left( 1 \right)$ and equation$\left( 2 \right)$ by putting value $n\text{ }=\text{ }23$

1st consecutive number $=\text{ }\dfrac{{{n}^{2}}\text{ }-\text{ }1}{2}$
$=\text{ }\dfrac{{{23}^{2}}\text{ }-\text{ }1}{2}$
$=\text{ }\dfrac{529\text{ }-\text{ }1}{2}$
$=\text{ }\dfrac{528}{2}$
$=\text{ }264$

2nd consecutive number $=\text{ }\dfrac{{{n}^{2}}\text{ }+\text{ }1}{2}$
$=\text{ }\dfrac{{{23}^{2}}\text{ }+\text{ }1}{2}$
$=\text{ }\dfrac{529\text{ }+\text{ }1}{2}$
$=\text{ }\dfrac{530}{2}$
$=\text{ }265$
So the two consecutive numbers are 264 and 265.