Question

The value of $(101{)}^2$ is

• A. 10,200
• B. 10,201
• C. 10,202
• D. 10,203

Answer 1

The correct option is B 10,201

$\underset{–}{ConceptInvolved}$:
$(a+b{)}^2=a^2+2ab+b^2$

$\underset{–}{Method}$:
Using the FOIL rule.
F=First terms, O= Outside terms, I = Inside terms, L= Last terms.

$(101{)}^2=(101\left)\right(101)=(100+1{)}^2$
$(100+1)(100+1)=\left(100\right)(100+1)+\left(1\right)(100+1)$
$=\left(100\right)\left(100\right)+\left(100\right)\left(1\right)+\left(1\right)\left(100\right)+(1{)}^2$
$=(100{)}^2+2(1\left)\right(100)+1^2$
$=10,000+200+1$
$=10,201$

This is the required expression obtained using FOIL rule.
Also, we can use the identity
$(a+b{)}^2=a^2+2ab+b^2$.

$(101{)}^2=(100+1{)}^2$
$=(100{)}^2+2(1\left)\right(100)+1^2$
$=10,000+200+1$
$=10,201$