In the expansion of (x + a)^n,the sum of odd | vedclass

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(B) We know that the binomial expansion is given by: $(x + a)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 + \binom{n}{3}x^{n-3

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$(x^2 - a^2)^n$

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(B) We know that the binomial expansion is given by: $(x + a)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 + \binom{n}{3}x^{n-3}a^3 + \dots$ Let $P$ be the sum of odd-positioned terms and $Q$ be the sum of even-positioned terms: $P = \binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \dots$ $Q = \binom{n}{1}x^{n-1}a + \binom{n}{3}x^{n-3}a^3 + \dots$ Thus,$(x + a)^n = P + Q$. Similarly,for $(x - a)^n$: $(x - a)^n = \binom{n}{0}x^n - \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 - \binom{n}{3}x^{n-3}a^3 + \dots$ $(x - a)^n = P - Q$. Now,calculating $P^2 - Q^2$: $P^2 - Q^2 = (P + Q)(P - Q)$ $P^2 - Q^2 = (x + a)^n (x - a)^n$ $P^2 - Q^2 = ((x + a)(x - a))^n$ $P^2 - Q^2 = (x^2 - a^2)^n$.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

In the expansion of $(x + a)^n$,the sum of odd terms is $P$ and the sum of even terms is $Q$. Then the value of $(P^2 - Q^2)$ is:

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