If n is a positive integer, then the middle term in the expansion of (1+x)2 n is

Question

If n is a positive integer, then the middle term in the expansion of (1+x)2 n is (A) (2 ⋅ 4 ⋅ 6 ldots ldots ldots ldots .2 n/n !) 2 n xn (B) (1 ⋅ 3 ⋅ 5 ldots ldots ldots ldots ldots(2 n-1)/n !) xn (C) (3 ⋅ 5 ⋅ 7 ldots ldots ldots ldots(2 n+1)/n !) 2 n xn (D) (1 ⋅ 3 ⋅ 5 ldots ldots ldots(2 n-1)/n !) 2n xn

Tardigrade Admin · Accepted Answer

As 2 n is even, the middle term of the expansion (1+x)2 n is ((2 n/2)+1)t h, i.e., (n+1)t h term which is given by Tn+1 = 2 n Cn(1)2 n-n(x)n= 2 n Cn xn=((2 n) !/n ! n !) xn =(2 n(2 n-1)(2 n-2) ldots 4 ⋅ 3 ⋅ 2 ⋅ 1/n ! n !) xn =(1 ⋅ 2 ⋅ 3 ⋅ 4 ldots(2 n-2)(2 n-1)(2 n)/n ! n !) xn =([1 ⋅ 3 ⋅ 5 ldots(2 n-1)][2 ⋅ 4 ⋅ 6 ldots(2 n)]/n ! n !) xn =([1 ⋅ 3 ⋅ 5 ldots(2 n-1)] 2n[1 ⋅ 2 ⋅ 3 ldots n]/n ! n !) xn =([1 ⋅ 3 ⋅ 5 ldots(2 n-1)] n !/n ! n !) 2n xn =(1 ⋅ 3 ⋅ 5 ldots(2 n-1)/n !) 2n xn

Q. If $n$ is a positive integer, then the middle term in the expansion of $(1 + x)^{2 n}$ is

$\frac{2 \cdot 4 \cdot 6 \dots\dots\dots\dots .2 n}{n !} 2 n x^{n}$

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

$\frac{3 \cdot 5 \cdot 7 \dots\dots\dots\dots ( 2 n + 1 )}{n !} 2 n x^{n}$

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

Q. If n is a positive integer, then the middle term in the expansion of (1+x)2n is

n!2⋅4⋅6………….2n​2nxn

n!1⋅3⋅5……………(2n−1)​xn

n!3⋅5⋅7…………(2n+1)​2nxn

n!1⋅3⋅5………(2n−1)​2nxn

Q. If $n$ is a positive integer, then the middle term in the expansion of $(1 + x)^{2 n}$ is

$\frac{2 \cdot 4 \cdot 6 \dots\dots\dots\dots .2 n}{n !} 2 n x^{n}$

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

$\frac{3 \cdot 5 \cdot 7 \dots\dots\dots\dots ( 2 n + 1 )}{n !} 2 n x^{n}$

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