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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2015 · Mathematics · Binomial Theorem

Consider the expansion of (1+x)^ 2 n+1 If the coefficients of x^r...

Consider the expansion of (1+x)2n+1(1+x)^{2 n+1} If the coefficients of xrx^r and xr+1x^{r+1} are equal in the expansion, then rr is equal to

Options

  1. An
  2. B2n−12\frac{2 \mathrm{n}-1}{2}
  3. C2n+12\frac{2 \mathrm{n}+1}{2}
  4. Dn+1\mathrm{n}+1

Correct answer

A. n

Step-by-step solution

(1+x)2n+1=(2n+1)C0x0+(2n+1)C1x1+….+((2n+1)2n+1)(x)2n+1\begin{aligned} & (1+x)^{2 n+1}={ }^{(2 n+1)} C_0 x^0+{ }^{(2 n+1)} C_1 x^1 \\ & \quad+\ldots .+\binom{(2 n+1)}{2 n+1}(x)^{2 n+1} \end{aligned} Coefficient of \mathrmxr=(2n+1)Cr\mathrmx^{\mathrm{r}}={ }^{(2 \mathrm{n}+1)} \mathrm{Cr} Coefficient of \mathrmxr+1=(2n+1)Cr+1\mathrmx^{\mathrm{r}+1}={ }^{(2 \mathrm{n}+1)} \mathrm{Cr}+1(2n+1)Cr=(2n+1)Cr+1⇒(2n+1)!r!(2n+1−r)!=(2n+1)!(r+1)!(2n−r)!⇒(2n−r)!(2n+1−r)(2n−r)!=r!(r+1)r!⇒(r+1)=2n+1−r⇒r=n\begin{aligned} & { }^{(2 n+1)} C_r={ }^{(2 n+1)} \mathrm{Cr}+1 \\ & \Rightarrow \frac{(2 n+1)!}{r!(2 n+1-r)!}=\frac{(2 n+1)!}{(r+1)!(2 n-r)!} \\ & \Rightarrow \frac{(2 n-r)!}{(2 n+1-r)(2 n-r)!}=\frac{r!}{(r+1) r!} \\ & \Rightarrow(r+1)=2 n+1-r \\ & \Rightarrow r=n \end{aligned}$$
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