Question

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

• A. $4(A+B)$
• B. $4(A-B)$
• C. $AB$
• D. $4AB$

Answer 1

The correct option is D $4AB$

$A$=sum of odd terms
Let n is odd,let total $T_{n+1}$ terms
$A=T_1+T_3+...T_n$
$A{=}^n{C}_0{+}^n{C}_2+....{+}^n{C}_{n-1}$
$A=\frac{2^n}{2}=2^{n-1}$
similarly $B=T_2+T_4+...T_{n+1}$
$B{=}^n{C}_1{+}^n{C}_3+....{+}^n{C}_n$
$B=\frac{2^n}{2}=2^{n-1}$
$A+B=2^n$
$(x+a{)}^{2n}-(x-a{)}^{2}n\Rightarrow (x+a+x-a{)}^n(x+a-x+a{)}^n$
$\left(2x{)}^n\right(2a{)}^n\Rightarrow 2^{2n}{x}^n{a}^n=4^n{a}^n{x}^n$
Considering this term being $4^n$
$AB=\frac{2^{2n}}{4}$
$4AB=2^{2n}=4^n$
Therefore,$(x+a{)}^{2n}-(x-a{)}^{2n}=4AB$