Question

If 'a' be the sum of the odd terms & b be the sum of the even terms in the expansion of ${(1+x)}^n$ then ${(1-x^2)}^n$

• A. $a^2-b^2$
• B. $a^2+b^2$
• C. $b^2-a^2$
• D. none

Answer 1

The correct option is A $a^2-b^2$

Sum of odd terms in the expansion of $(1+x{)}^n$ will be $2^{n-1}$ ...(i)
Now
$(1-x^2{)}^n={}^n{C}_0-{}^n{C}_1{x}^2+{}^n{C}_2{x}^4+...(-1{)}^n{}^n{C}_n{x}^{2n}$ ...(n)
$(1+x^2{)}^n={}^n{C}_0+{}^n{C}_1{x}^2+{}^n{C}_2{x}^4+...{}^n{C}_n{x}^{2n}$ ...(m)
Subtracting n from m, we get
$(1+x^2{)}^n-(1-x^2{)}^n=2[{}^n{C}_1{x}^2+{}^n{C}_3{x}^6+...]$
Now let x=1.
Hence we get
$2^n-0=2[{}^n{C}_1+{}^n{C}_3+...]$
Or
${}^n{C}_1+{}^n{C}_3{x}^2+{}^n{C}_5+...=2^{n-1}$
Hence
$a=b$
Now
$(1-x^2{)}^n$
$=(1+x{)}^n(1-x{)}^n$
$\to \left[\right(a+b\left)\right(a-b\left)\right]$
$=a^2-b^2$.