Question

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $(x+a{)}^n$ are P and Q respectively then the value of $(P^2-Q^2)$ equal to

• A. $(x^2-a^2{)}^{2n}$
• B. $(x^2+a^2{)}^{2n}$
• C. $(x+a{)}^n$
• D. $(x^2-a^2{)}^n$

Answer 1

The correct option is D $(x^2-a^2{)}^n$

$(x+a{)}^n{=}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}a{+}^n{C}_2{x}^{n-2}{a}^2{+}^n{C}_3{x}^{n-3}{a}^3+.....$
$=(x^n{+}^n{C}_2{x}^{n-2}{a}^2+.....)+{(}^n{C}_1{x}^{n-1}a{+}^n{C}_3{x}^{n-3}{a}^3+....)$
$\Rightarrow (x+a{)}^n=P+Q$ ......(1)
Now, $(x-a{)}^n{=}^n{C}_0{x}^n{-}^n{C}_1{x}^{n-1}a{+}^n{C}_2{x}^{n-2}{a}^2{-}^n{C}_3{x}^{n-3}{a}^3+.....$
$=(x^n{+}^n{C}_2{x}^{n-2}{a}^2+.....)-{(}^n{C}_1{x}^{n-1}a{+}^n{C}_3{x}^{n-3}{a}^3+....)$
$\Rightarrow (x-a{)}^n=P-Q$ .....(2)
Now, $P^2-Q^2=(P+Q)(P-Q)$
$=(x+a{)}^n(x-a{)}^n$
$\Rightarrow P^2-Q^2=(x^2-a^2{)}^n$