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

Explanation

Let,`(x+a)^n=n_(C_0)​x^n+n_(C_1)​x^(n−1)a^1+...+n_(C_n)​a^n`

=`(n_(C_0)​x^n+n_(C_2)​x^(n−2)a^2+...)+(n_(C_1)​x^(n−1)a^1+...)`=`A+B`

`(x-a)^n=n_(C_0)​x^n-n_(C_1)​x^(n−1)a^1+...-n_(C_n)​a^n`

=`(n_(C_0)​x^n+n_(C_2)​x^(n−2)a^2+...)-(n_(C_1)​x^(n−1)a^1+...)`=A-B

∴`A^2−B^2=(A+B)(A−B)=(x+a)^n(x−a)^n=(x^2−a^2)^n`