Question

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

• A. $1$
• B. $2$
• C. $5$
• D. None of the above.

Answer 1

Answer: (D)

None of the above.

Let the 3 consecutive terms be ${}^n{C}_{r-1}$, ${}^n{C}_r$ and ${}^n{C}_{r+1}$.

So, $\frac{1}{{}^n{C}_{r-1}}+\frac{1}{{}^n{C}_{r+1}}=\frac{2}{{}^n{C}_r}\Rightarrow \frac{(n-r+1)!(r-1)!}{n!}+\frac{(n-r-1)!(r+1)!}{n!}=\frac{2(n-r)!\left(r\right)!}{n!}\Rightarrow \frac{(n-r+1)}{r}+\frac{(r+1)}{(n-r)}=2\Rightarrow {(n-r)}^2+(n-r)+r(r+1)=2(n-r)r$

On solving, $(n-2r{)}^2=0$ So, $n=2r$

Hence, max. value of $n$ can't be determined. So, $D$ is correct.