Question

$\frac12\left\{\left(a+b\right)^2+\left(a-b\right)^2\right\}$ = কত?

Options :

• (A) $a^2+b^2$
• (B) $\frac{\left(a+b\right)^2}2-\frac{\left(a-b\right)^2}2$
• (C) $a^2-b^2$
• (D) $\left(a+b\right)^2+\left(a-b\right)^2$

Answer 1

$\frac12\left\{\left(a+b\right)^2+\left(a-b\right)^2\right\}$

= $\frac12\left\{a^2+2ab+b^2+a^2-2ab+b^2\right\}$

= $\frac12\left\{a^2+\cancel{2ab}+b^2+a^2-\cancel{2ab}+b^2\right\}$

= $\frac12\left\{2a^2+2b^2\right\}$

= $\frac12.2\left\{a^2+b^2\right\}$

= $a^2+b^2$