Question

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

• A. ${(x^3-a^2)}^n$.
• B. ${(x^2-a^2)}^n$.
• C. ${(x^2-a^3)}^n$.
• D. ${(x^3-a^3)}^n$.
  

Answer 1

Answer: (B)

${(x^2-a^2)}^n$.

$(x+a{)}^n={}^n{C}_0{x}^n+{}^n{C}_{1}a(x{)}^{n-1}+\dots +{}^n{c}_n{a}^n$

$(x-a{)}^n=n{c}_0{x}^n-{}^n{c}_{1}a(x{)}^{n-1}+\cdots +(-1{)}^n{}^n{C}_n\cdot a^n$

$\text{Adding, we get}$

$A=(x+a{)}^n+(x-a{)}^n$

$=2({}^n{c}_0{x}^n+{}^n{c}_2{a}^2(x{)}^{n-2}\dots +)$

$\text{subtracting, we get}$

$B=(x+a{)}^n-(x-a{)}^n$

$=2\left(n{C}_{1}a\right(x{)}^{n-1}+{}^n{c}_3{a}^3(x{)}^{n-3}+\cdots )$

$\Rightarrow A=\frac{(x+a{)}^n+(x-a{)}^n}{2}$

$B=\frac{(x+a{)}^n-(x-a{)}^n}{2}$

$A^2-B^2=(A+B)(A-B)$

$=(x+a{)}^n(x-a{)}^n$

$\text{Using}(a+b)(a-b)=a^2-b^2$

$A^2-B^2={(x^2-a^2)}^n$