Question

If '$a$' be the sum of the odd terms and '$b$' the sum of the even terms of the expansion of $(1+x{)}^n$, then $(1-x^2{)}^n=$

• A. $a^2-b^2$
• B. $a^2+b^2$
• C. $b^2-a^2$
• D. $n(a+b)$

Answer 1

The correct option is A $a^2-b^2$

We know that ${(1+x)}^n=C_1+C_{2}x+...+C_n{x}^n=a+b$,

Replacing $x$ by $-x$ gives,

${(1-x)}^n=C_1-C_{2}x+...+{{(-1)}^{n}C}_n{x}^n⟹a-b$,

$\therefore {(1-x^2)}^n={(1+x)}^n{(1-x)}^n=(a+b)(a-b)=a^2-b^2$