Question

If $P$ be the sum of odd terms and $Q$ be the sum of even terms in the expansion of $(x+a{)}^n,$ then $P^2-Q^2=(x^2+a^2{)}^n$.

• A. True
• B. False

Answer 1

The correct option is B False

$(x+a{)}^n=({}^n{C}_0{x}^n{a}^0+{}^n{C}_2{x}^{n-2}{a}^2+....)+({}^n{C}_1{x}^{n-1}+{}^n{C}_3{x}^{n-3}{a}^3...)$

$\therefore$ $(x+a{)}^n=P+Q$ ----- ( 1 )

Also,

$(x-a{)}^n={}^n{C}_0{x}^n{a}^0-{}^n{C}_1{x}^{n-1}{a}^1+{}^n{C}_2{x}^{n-2}{a}^2-{}^n{C}_3{x}^{n-3}{a}^3+...+(-1{)}^n{}^n{C}_n{x}^0{a}^n$

$\therefore$ $(x-a{)}^n=({}^n{C}_0{x}^n{a}^0+{}^n{C}_2{x}^{n-2}{a}^2+....)-({}^n{C}_1{x}^{n-1}{a}^1+{}^n{C}_3{x}^{n-3}{a}^3)$

$\Rightarrow$ $(x-a{)}^n=P-Q$ ----- ( 2 )

$\Rightarrow$ $P^2-Q^2=(P+Q)(P-Q)$

$\Rightarrow$ $P^2-Q^2=(x+a{)}^n(x-a{)}^n$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $P^2-Q^2=(x^2-a^2{)}^n$

$\therefore$ The statement given in given question is false.