Question

If the coefficients of three consecutive terms in the expansion of ${\left(1+x\right)}^n$ are in the ratio $1:7:42,$ then the value of $n$ is equal to

• A. $49$
• B. $50$
• C. $55$
• D. $56$

Answer 1

Answer: (C)

For $T_{r+1}={}^n{C}_r{x}^r,$ coefficient $\to {}^n{C}_r$
${}^n{C}_{r-1}:{}^n{C}_r:{}^n{C}_{r+1}=1:7:42$
$\Rightarrow \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{7}{1}$ and $\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{42}{7}$
$\Rightarrow \frac{n-r+1}{r}=7$ and $\frac{n-r}{r+1}=6$
$\Rightarrow n-r+1=7r$ and $n-r=6r+6$
$\Rightarrow n=8r-1$ and $n=7r+6$
$\Rightarrow 8r-1=7r+6\Rightarrow r=7$
$\Rightarrow n=55$