Question

Algebric identity for (99) ²= (100+1)²

Answer 1

$99^{2} = (100+1)^{2}$

$= 100^{2} + 2\times 100 \times 1 + 1^{2}$

$\boxed { \pink { (a+b)^{2} = a^{2} + 2ab + b^{2} }}$

$= 10000 + 200 + 1\\= 10201$

Therefore.,

$\red{ 99^{2}} \green { = 10201}$

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