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  • If A is the sum of odd n terms and B is the sum of even terms in the expansion of , then

If A is the sum of odd n terms and B is the sum of even terms in the expansion of (x+a)^n, then A^2-B^2-k\left(x^2-a^2\right)^n=0, where k=

Option: 1

1


Option: 2

2


Option: 3

-1


Option: 4

3


Answers (1)

best_answer

\begin{aligned} (x+a)^n & ={ }^n C_0 x^n+{ }^n C_1 x^{n-1} a+{ }^n C_2 x^{n-2} a^2+\ldots . .+{ }^n C_n a^n \\ & =A+B \end{aligned}\\ $$ where\\ $$ \begin{gathered} A={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+{ }^n C_4 x^{n-4} a^4+\ldots . . \\ \quad B={ }^n C_1 x^{n-1}+{ }^n C_3 x^{n-3} a^3+{ }^n C_5 x^{n-5} a^5+\ldots . . \\ \therefore \quad(x-a)^n=A-B \\ A^2-B^2=(A-B)(A+B) \\ =(x-a)^n(x+a)^n \\ \quad=\left(x^2-a^2\right)^n \\ \Rightarrow \quad k=1 \end{gathered}

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Irshad Anwar

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