Question

If binomial coefficients of three consecutive terms of ${(1+x)}^n$ are in HP, then the maximum value of $n$ is

• A. $1$
• B. $2$
• C. $0$
• D. None of these

Answer 1

Answer: (D)

None of these

Let the coefficients of $r$th, $(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms be in HP

Then, $\frac{2}{{{}^{n}C}_r}=\frac{1}{{{}^{n}C}_{r-1}}+\frac{1}{{{}^{n}C}_{r+1}}$

$\Rightarrow 2=\frac{{{}^{n}C}_r}{{{}^{n}C}_{r-1}}+\frac{{{}^{n}C}_r}{{{}^{n}C}_{r+1}}$

$\Rightarrow 2=\frac{n-r+1}{r}+\frac{r+1}{n-r}$

$\Rightarrow n^2-4nr+4r^2+n=0$

$\Rightarrow {(n-2r)}^2+n=0$

which is not possible for any value for $n$