The middle term in the expansion of (1 + x)^{ | vedclass

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Question

(D) The expansion of $(1 + x)^{2n}$ has $2n + 1$ terms,which is an odd number. Therefore,the middle term is the $(n + 1)$-th term,denoted as $T_{n+1}$

Vedclass · Accepted Answer

$\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!} 2^n x^n$

Vedclass · Answer

(D) The expansion of $(1 + x)^{2n}$ has $2n + 1$ terms,which is an odd number. Therefore,the middle term is the $(n + 1)$-th term,denoted as $T_{n+1}$. Using the general term formula $T_{r+1} = {}^{2n}C_r x^r$,we get $T_{n+1} = {}^{2n}C_n x^n$. Expanding the binomial coefficient: ${}^{2n}C_n = \frac{(2n)!}{n! n!} = \frac{[1 \cdot 2 \cdot 3 \cdot \dots \cdot (2n)]}{n! n!}$. Separating the odd and even terms in the numerator: ${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot [2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)]}{n! n!}$. Factoring out $2$ from each of the $n$ even terms: ${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot 2^n \cdot [1 \cdot 2 \cdot 3 \cdot \dots \cdot n]}{n! n!}$. Since $1 \cdot 2 \cdot 3 \cdot \dots \cdot n = n!$,we have: ${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot 2^n \cdot n!}{n! n!} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!} 2^n$. Thus,the middle term is $\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!} 2^n x^n$.

The middle term in the expansion of $(1 + x)^{2n}$ is

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