7.Binomial Theorem
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दिखाइए कि $(1+x)^{2 n}$ के प्रसार में मध्य पद का गुणांक, $(1+x)^{2 n-1}$ के प्रसार में दोनों मध्य पदों के गुणांकों के योग के बराबर होता है।

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

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