Question

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

• A. 95
• B. 55
• C. 35
• D. 11

Answer 1

The correct option is B 55

Let the three consecutive terms be $(r-1)th,rth,(r+1)th$ terms.

That is, $T_{r-1},T_r.T_{r+1}$.

We know that general term of expansion $(a+b{)}^n$ is

$T_{r+1}={n_C}_r{a}^{n-r}{b}^r$

For $(1+x{)}^n$,

Putting $a=1,b=x$,

$T_{r+1}={n_C}_r{1}^{n-r}{x}^r$

$T_{r+1}={n_C}_r{x}^r$

Therefore, coefficient of $(r+1)th$ term = ${n_C}_r$

For $rth$ term of $(1+x{)}^n$

Replacing r with r-1, we get,

$T_r={n_C}_{r-1}{x}^{r-1}$

Therefore, coefficient of $rth$ term = ${n_C}_{r-1}$

For $(r-1)th$ term of $(1+x{)}^n$

Replacing r with r-2. we get,

$T_{r-1}={n_C}_{r-2}{x}^{r-2}$

Therefore, coefficient of $(r-1)th$ term = ${n_C}_{r-2}$

Since the coefficient of (r-1)th, rth and (r+1)th terms are in the ratio 1 : 7 : 42,

$\frac{{n_C}_{r-2}}{{n_C}_{r-1}}=\frac{1}{7}$

$\frac{\frac{n!}{(r-2)![n-(r-2\left)\right]!}}{\frac{n!}{(r-1)![n-(r-1\left)\right]!}}=\frac{1}{7}$

$\frac{(r-1)[n-(r-1\left)\right]!}{[n-(r-2\left)\right]!}=\frac{1}{7}$

$\frac{(r-1)(n-r+1)!}{(n-r+2)!}=\frac{1}{7}$

$\frac{(r-1)(n-r+1)!}{(n-r+2)(n-r+1)!}=\frac{1}{7}$

$\frac{(r-1)}{(n-r+2)}=\frac{1}{7}$

$n-8r+9=0$ ......(1)

Also,

$\frac{\frac{n!}{(r-1)![n-(r-1\left)\right]!}}{\frac{n!}{r!(n-r)!}}=\frac{7}{42}$

$\frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}=\frac{1}{6}$

$\frac{r(n-r)!}{(n-r+1)!}=\frac{1}{6}$

$\frac{r}{n+1-r}=\frac{1}{6}$

$n-7r+1=0$ .....(2)

Solving equations 1 & 2, we get,

$r=8$ and $n=55$

Hence, $n=55$