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

Solution

1. Let the three consecutive coefficients be $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$, $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, and $\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)$, which are in the ratio $1:7:42$.
2. From the ratio $\frac{\left(\genfrac{}{}{0ex}{}{n}{r}\right)}{\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)}=7$, we obtain $\frac{n-r+1}{r}=7$, which simplifies to $n-r+1=7r$, or $n=8r-1$.
3. From the ratio $$\begin{array}{rl}\frac{\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{n}{r}\right)}& =\frac{42}{7}\\ & =6\end{array}$$, we obtain $\frac{n-r}{r+1}=6$, which simplifies to $n-r=6r+6$, or $n=7r+6$.
4. Equating the two expressions for $n$, we have $8r-1 = 7r+6$, which gives $r=7$.
5. Substituting $r=7$ back into $n=8r-1$, we find $$\begin{array}{rl}n& =8(7)-1\\ & =55\end{array}$$.
6. Hence the answer is (A).