Question

express the following as the sum of two consecutive natural numbers are 97 square​

Answer 1

☆Answer☆

• $\sf {{97}^{2} }$= 4704 + 4705

Step-by-step explanation-

We have to use given formula-

$\huge\rm\red {{n}^{2} = \frac{ {n}^{2} - 1}{2} + \frac{ {n}^{2} + 1 }{2}}$

• ${{n}^{2}}$ = given natural number ( 97^2)

Now, putting values

$\huge\sf\blue { {97}^{2} = \frac{ {97}^{2} - 1}{2} + \frac{ {97}^{2} + 1 }{2}}$

$\sf {{97}^{2} = \frac{9409 - 1}{2} + \frac{9409 + 1}{2}}$

$\sf { {97}^{2} = \frac{9408}{2} + \frac{9500}{2}}$

$\sf\green { {97}^{2} = 4704 + 4705 \: ans..}$

Hence, $\sf {{97}^{2} }$ is the sum of two consecutive natural numbers which are 4704 and 4705.