Question

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

• A. $60$
• B. $55$
• C. $70$
• D. none of the above

Answer 1

Answer: (B)

$55$

Let the three consecutive terms be $(r-1{)}^{th},r^{th},(r+1{)}^{th}$ i.e $T_{r-1},T_r,T_{r+1}$

We know that feneral term of expansion $(1+x{)}^n$ is

$T_{r+1}={}^n{C}_r{x}^r----\left(1\right)$

$\therefore$ Coefficient of $(r+1{)}^{th}$ term $={}^n{C}_r$

For $r^{th}$ term replace r with $r-1$ in eq(1)

$\therefore$ Coefficient of $(r{)}^{th}$ term $={}^n{C}_{r-1}$

For $(r-1{)}^{th}$ term replace r with $r-2$ in eq(1)

$\therefore$ Coefficient of $(r-1{)}^{th}$ term $={}^n{C}_{r-2}$

Since the coefficient of $(r-1{)}^{th},r^{th},(r+1{)}^{th}$ terms are in ratio $1:7:42$

$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{7}$

$\frac{{}^n{C}_{r-1}}{{}^n{C}_{r-2}}=7$

We know that

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

Hence

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

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

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

$n-8r+9=0----\left(2\right)$

And

$\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{7}{42}$

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{42}{7}$

We know that

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

Hence

$\frac{n-r+1}{r-1}=\frac{42}{7}$

$7(n-r+1)=42(r-1)$

$7n-7r+7=42r-42$

$n-7r+1=0----\left(3\right)$

On solving eq (2) and (3) we get

$8r-9-7r+1=0$

$8r-9-7r+1=0$

$r=8$

Putting value of r in eq (2)

$n-8r+9=0$

$n-64+9=0$

$n=55$