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Question: Answered & Verified by Expert
Consider the expansion of $(1+\mathrm{x})^{2 \mathrm{n}+1}$
If the coefficients of $\mathrm{x}^{\mathrm{r}}$ and $\mathrm{x}^{\mathrm{r}+1}$ are equal in the expansion, then $\mathrm{r}$ is equal to
MathematicsBinomial TheoremNDANDA 2015 (Phase 2)
Options:
  • A $\mathrm{n}$
  • B $\frac{2 \mathrm{n}-1}{2}$
  • C $\frac{2 \mathrm{n}+1}{2}$
  • D $\mathrm{n}+1$
Solution:
1131 Upvotes Verified Answer
The correct answer is: $\mathrm{n}$
$(1+\mathrm{x})^{2 \mathrm{n}+1}={ }^{(2 \mathrm{n}+1)} \mathrm{C}_{0} \mathrm{x}^{0}+{ }^{(2 \mathrm{n}+1)} \mathrm{C}_{1} \mathrm{x}^{1}$
$+\ldots+{ }^{(2 n+1)} C_{2 n+1}(x)^{2 n+1}$
Coefficient of $\mathrm{x}^{\mathrm{r}}=(2 \mathrm{n}+1) \mathrm{Cr}$
Coefficient of $\mathrm{x}^{\mathrm{r}+1}=(2 \mathrm{n}+1) \mathrm{Cr}+1$
$(2 \mathrm{n}+1) \mathrm{C}_{\mathrm{r}}$=$(2 n+1) C r+1$
$\Rightarrow \frac{(2 \mathrm{n}+1) !}{\mathrm{r} !(2 \mathrm{n}+1-\mathrm{r}) !}=\frac{(2 \mathrm{n}+1) !}{(\mathrm{r}+1) !(2 \mathrm{n}-\mathrm{r}) !}$
$\Rightarrow \frac{(2 \mathrm{n}-\mathrm{r}) !}{(2 \mathrm{n}+1-\mathrm{r})(2 \mathrm{n}-\mathrm{r}) !}=\frac{\mathrm{r} !}{(\mathrm{r}+1) \mathrm{r} !}$
$\Rightarrow(\mathrm{r}+1)=2 \mathrm{n}+1-\mathrm{r}$
$\Rightarrow \mathrm{r}=\mathrm{n}$

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