Question

the coefficients of three consecutive terms. in the expansion of (1 + a)n are in the ratio 1: 7 :42 find n​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer: n= 55

Step-by-step explanation:

Let the three consecutive terms in the expansion $(1+a)^n$

are $(r-1)^{th} , r^{th} , (r+1)^{th}$

The coefficient of  $(r-1)^{th}$ is $_{r-2}^{n}\textrm{C}$

The coefficient of  $(r)^{th}$ is $_{r-1}^{n}\textrm{C}$

The coefficient of  $(r+1)^{th}$ is $_{r}^{n}\textrm{C}$

Now coefficients are in ratio 1:7:42

therefore

$\dfrac{_{r-2}^{n}\textrm{C}}{_{r-1}^{n}\textrm{C}} = \dfrac{1}{7} \\\\\Rightarrow \dfrac{\dfrac{n!}{(r-2)!(n-(r-2))!} }{\dfrac{n!}{r!(n-(r-1))!} } =\dfrac{1}{7} \\\\\Rightarrow n-8r+9= 0$

Similarly

$\dfrac{_{r-1}^{n}\textrm{C}}{_{r}^{n}\textrm{C}} = \dfrac{7}{42} \\\\\Rightarrow \dfrac{\dfrac{n!}{(r-1)!(n-(r-1))!} }{\dfrac{n!}{r!(n-r)!} } =\dfrac{7}{42} \\\\\Rightarrow n-7r+1= 0$

Solving the equation we get n= 55

Hence the value of n is 55