Question

If $A$ is the sum of the odd terms and $B$ the sum of even terms in the expansion of ${(x+a)}^n$, then $A^2-B^2=$

• A. ${(x^2+a^2)}^n$
• B. ${(x^2-a^2)}^n$
• C. $2{(x^2-a^2)}^n$
• D. None of these

Answer 1

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

We have,

${(x+a)}^n{=}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}{a}^1{+}^n{C}_2{x}^{n-2}{a}^2{+}^n{C}_3{x}^{n-3}{a}^3+...{+}^n{C}_n{a}^n=({}^n{C}_0{x}^n{+}^n{C}_2{x}^{n-2}{a}^2+...)+({}^n{C}_1{x}^{n-1}{a}^1{+}^n{C}_3{x}^{n-3}{a}^3+...)=A+B{(x-a)}^n{=}^n{C}_0{x}^n{-}^n{C}_1{x}^{n-1}{a}^1{+}^n{C}_2{x}^{n-2}{a}^2{-}^n{C}_3{x}^{n-3}{a}^3=({}^n{C}_0{x}^n{+}^n{C}_2{x}^{n-2}{a}^2+...)-({}^n{C}_1{x}^{n-1}{a}^1{+}^n{C}_3{x}^{n-3}{a}^3+...)=A-B\therefore A^2-B^2=(A+B)(A-B)={(x+a)}^n{(x-a)}^n={(x^2-a^2)}^n$