The sum of the coefficients of the middle terms of (1+x)^ 2 n-1 .… - Vidyadip

MCQ

The sum of the coefficients of the middle terms of $\left(1+x)^{2 n-1}\right.$ is:

A
${ }^{2 n-1} C_n$
B
${ }^{2 n-1} C_{n+1}$
C
${ }^{2 n} \mathrm{C}_{\mathrm{n}-1}$
✓
${ }^{2 n} \mathrm{C}_n$

✓

Answer

Correct option: D.

${ }^{2 n} \mathrm{C}_n$

${ }^{2 n} \mathrm{C}_n$

Solution
Consider: $(1+x)^{2 n-1}$
Since $2 \mathrm{n}-1$ is odd, the middle terms are $\left(\frac{2 \mathrm{n}-1+1}{2}\right)^{\text {th }}$ and $\left(\frac{2 \mathrm{n}-1+1}{2}+1\right)^{\text {th }}$ terms
Now, consider the following
$T r+1={ }^n C_r a^{n-r} b^r \ldots \ldots(i)$
Where T represents the term
Here the middle terms are the $n$th term and $(n+1)^{\text {th }}$ term.
$T_{n+1}={ }^{2 n-1} C_n 1^{n-1} x^n={ }^{2 n-1} C_n x^n$
Hence, the coefficient is ${ }^{2 n-1} C_n \ldots$ (1)
$T_n={ }^{2 n-1} C_{n-1} 1^n x^{n-1}={ }^{2 n-1} C_{n-1} x^{n-1}$
Hence, ${ }^{2 n-1} C_{n-1}$ is the coefficient ...(2)
Therefore, sum of the coefficients of the middle terms is
${ }^{2 n-1} C_n+{ }^{2 n-1} C_{n-1}$
$={ }^{2 n} C_n$

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