The middle term in the expansion of (1 + x)2n is

Question

The middle term in the expansion of (1 + x)2n is (A) (1⋅ 3⋅ 5....(2n-1)/n) xn (B) (1⋅ 3⋅ 5....(2n-1)/n!) 2n-1 xn (C) (1⋅ 3⋅ 5....(2n-1)/n!) xn (D) (1⋅ 3⋅ 5....(2n-1)/n!) 2n xn

Tardigrade Admin · Accepted Answer

Total term = (2n + 1) Middle term = ((2n+1) +1/2) = (n +1)th term Tn+1 = 2nCn xn =(2n!/n!n!) xn =(2n n! 1⋅ 3⋅ 5..... (2n-1) /n!n!)xn = (1⋅ 3⋅ 5 .....(2n-1)/n!) 2n xn

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

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

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

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

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

Total term = $(2 n + 1)$
Middle term = $\frac{( 2 n + 1 ) + 1}{2} = (n + 1)^{t h}$ term
$T_{n + 1} =^{2 n} C_{n} x^{n}$
$= \frac{2 n !}{n ! n !} x^{n}$
$= \frac{2 ^{n} n ! { 1 \cdot 3 \cdot 5..... ( 2 n - 1 ) }}{n ! n !} x^{n}$
$= {\frac{1 \cdot 3 \cdot 5..... ( 2 n - 1 )}{n !}} 2^{n} x^{n}$

Q. The middle term in the expansion of (1+x)2n is

n1⋅3⋅5....(2n−1)​xn

n!1⋅3⋅5....(2n−1)​2n−1xn

n!1⋅3⋅5....(2n−1)​xn

n!1⋅3⋅5....(2n−1)​2nxn