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

^{2 n - 1} C_{n}

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^{2 n - 1} C_{n + 1}

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^{2 n} C_{n - 1}

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^{2 n} C_{n}

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Solution

The correct option is C $^{2 n} C_{n}$
Consider: $(1 + x)^{2 n - 1}$
Since $2 n - 1$ is odd, the middle terms are ${(\frac{2 n - 1 + 1}{2})}^{t h}$ and ${(\frac{2 n - 1 + 1}{2} + 1)}^{t h}$ terms
Now, consider the following
$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ ...(i)
Where $T$ represents the term
Here the middle terms are the $n^{t h}$ term and $(n + 1)^{t h}$ 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 $^{2 n - 1} C_{n}$ ...(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
$^{2 n - 1} C_{n} +^{2 n - 1} C_{n - 1}$
$=^{2 n} C_{n}$

Hence, option D.

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Similar questions

1. {C_{0}}^{2} + 3. {C_{1}}^{2} + 5. {C_{2}}^{2} + . . . + (2 n + 1) . {C_{n}}^{2} = 2 n .^{2 n - 1} C_{n} +^{2 n} C_{n} .

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

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

^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + . . . . . . +^{2 n + 1} C_{n} +^{2 n + 1} C_{n + 1} +^{2 n + 1}

C_{n + 2} + . . . . . +^{2 n + 1} C_{2 n + 1}

n = 3

\sum_{r = 0}^{n} r (^{n} C_{r})^{2} = n (^{2 n - 1} C_{n - 1})

\sum_{r = 0}^{n} (^{n} C_{r})^{2} =^{2 n} C_{n}

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