Show that coefficient of middle term in expansion of (1+x)^{2 n} is equal to sum of coefficients of

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7.Binomial Theorem

medium

Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$

Option A

Option B

Option C

Option D

Solution

As $2 n$ is even so the expansion $(1+x)^{2 n}$ has only one middle term which is
$\left(\frac{2 n}{2}+1\right)^{\text {th }}$ i.e., $(n+1)^{\text {th }}$ term.

The $(n+1)^{\text {th }}$ term is $^{2 n} C_{n} x^{n}$. The coefficient of $x^{n}$ is $^{2 n} C_{n}$

Similarly, $(2 n-1)$ being odd, the other expansion has two middle terms,

$\left(\frac{2 n-1+1}{2}\right)^{ th }$ and $\left(\frac{2 n-1+1}{2}+1\right)^{ th }$ i.e., $n^{ th }$ and $(n+1)^{ th }$ terms. The coefficients of these terms are $^{2n – 1}{C_{n – 1}}$ and $^{2n – 1}{C_n},$ respectively.

$^{2n – 1}{C_{n – 1}} + {\,^{2n – 1}}{C_n} = {\,^{2n}}{C_n}$ [ As ${^n{C_{r – 1}} + {\,^n}{C_r} = {\,^{n + 1}}{C_r}}$ ] as required.

Standard 11

Mathematics

If coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the binomial expansion of ${(1 + x)^n}$ are in $A.P.$, then ${n^2} – 9n$ is equal to

medium

If the coefficient of $x ^7$ in $\left(a x-\frac{1}{b x^2}\right)^{13}$ and the coefficient of $x^{-5}$ in $\left(a x+\frac{1}{b x^2}\right)^{13}$ are equal, then $a^4 b^4$ is equal to :

hard

(JEE MAIN-2023)

Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$

Solution

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Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$

Solution

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