Question

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

• A. $A^2-B^2$
• B. $A^2+B^2$
• C. $4AB$
• D. None

Answer 1

Answer: (A)

$(x+a{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}a+\dots \dots \dots \dots \dots$
$=\left({}^n{C}_0{x}^n+{}^n{C}_2{x}^{n-2}{a}^2+\dots \dots \dots \right)$
$+\left({}^n{C}_1{x}^{n-1}a+{}^n{C}_3{x}^{n-3}{a}^3+\dots \dots \right)$
Given $:A={}^n{C}_0{x}^n+{}^n{C}_2{x}^{n-2}{a}^2+\dots \dots \dots \dots \dots$
$B={}^n{C}_1{x}^{n-1}\cdot a+{}^n{C}_3{x}^{n-3}{a}^3+\dots \dots \dots \dots$
$\Rightarrow (x+a{)}^n=A+B$
$(x-a{)}^n={}^n{C}_0{x}^n-{}^n{C}_1{x}^{n-1}a+{}^n{C}_2{x}^{n-2}{a}^2-\dots \dots \dots .$
$=\left({}^n{C}_0{x}^n+{}^n{C}_2{x}^{n-2}{a}^2+\dots \dots \dots \right)$
$-\left({}^n{C}_1{x}^{n-1}a+{}^n{C}_3{x}^{n-3}{a}^3+\dots \right)$
$\Rightarrow (x-a{)}^n=A-B$
$\therefore {\left(x^2-a^2\right)}^n=(x+a{)}^n\cdot (x-a{)}^n$
$=(A+B)(A-B)=A^2-B^2$