Question

If a, b, c be the three consecutive coefficients in the expansion of a power of ${{\left( 1+x \right)}^{n}}$ , then $n=?$
$\begin{align}
  & \text{A}\text{. }\dfrac{2ac+b\left( a+c \right)}{{{b}^{2}}-ac} \\
 & \text{B}\text{. a+b-2c} \\
 & \text{C}\text{. }\dfrac{2ac}{{{b}^{2}}-ac} \\
 & \text{D}\text{. None of these}\text{.} \\
\end{align}$

Answer 1

Hint: To find this question first we need to consider $a={}^{n}{{C}_{r}},b={}^{n}{{C}_{r+1}},c={}^{n}{{C}_{r+2}}$ by assuming $r,r+1\And r+2$ as the three consecutive coefficients. Then find the value of ‘r’ by dividing $\dfrac{a}{b}$ & $\dfrac{b}{c}$ and equalize them to find the value of ‘n’.

Complete step by step solution:
Let us consider the consecutive coefficients to be $r,r+1,r+2$ terms.
So, we will consider a, b, c as –
$\begin{align}
  & a={}^{n}{{C}_{r}} \\
 & b={}^{n}{{C}_{r+1}} \\
 & c={}^{n}{{C}_{r+2}} \\
\end{align}$
Here, we will find the value of ‘r’ by dividing a by b we get –
$\dfrac{a}{b}=\dfrac{{}^{n}{{C}_{r}}}{{}^{n}{{C}_{r+1}}}$
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . So, we get –
$\dfrac{a}{b}=\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}}$
We can also write it as –
\[\dfrac{a}{b}=\dfrac{n!}{r!\left( n-r \right)!}\times \dfrac{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}{n!}\]
By simplifying this we get –
\[\dfrac{a}{b}=\dfrac{n!}{r!\left( n-r \right)!}\times \dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{n!}\]
By cancelling the common factors from numerator and denominator, we get –
$\dfrac{a}{b}=\dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{r!\left( n-r \right)!}$
Here, $\left( r+1 \right)!$ can be written as $\left( r+1 \right)r!$ and $\left( n-r \right)!$ can be written as $\left( n-r \right)\left( n-r-1 \right)!$
So we get –
$\dfrac{a}{b}=\dfrac{\left( r+1 \right)r!\left( n-r-1 \right)!}{r!\left( n-r \right)\left( n-r-1 \right)!}$
By cancelling the common factors from numerator and denominator, we get –
$\dfrac{a}{b}=\dfrac{r+1}{n-r}$
By cross multiplication, we get –
$\begin{align}
  & a\left( n-r \right)=b\left( r+1 \right) \\
 & an-ar=br+b \\
\end{align}$
By subtracting ‘b’ and adding ‘ar’ on both sides, we get –
$\begin{align}
  & an-b=br+ar \\
 & an-b=r\left( a+b \right) \\
\end{align}$
By dividing $a+b$ on the both sides, we get –
$r=\dfrac{an-b}{a+b}$ …………………….. (1)
Now we will divide b and c so we get an another value of ‘r’, so we get –
$\dfrac{b}{c}=\dfrac{{}^{n}{{C}_{r+1}}}{{}^{n}{{C}_{r+2}}}$
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . So, we get –
$\dfrac{b}{c}=\dfrac{\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}}{\dfrac{n!}{\left( r+2 \right)!\left( n-\left( r+2 \right) \right)!}}$
We can also write it as –
\[\dfrac{b}{c}=\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}\times \dfrac{\left( r+2 \right)!\left( n-\left( r+2 \right) \right)!}{n!}\]
By simplifying this we get –
\[\dfrac{b}{c}=\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}\times \dfrac{\left( r+2 \right)!\left( n-r-2 \right)!}{n!}\]
By cancelling the common factors from numerator and denominator, we get –
$\dfrac{b}{c}=\dfrac{\left( r+2 \right)!\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-1 \right)!}$
Here, $\left( r+2 \right)!$ can be written as $\left( r+2 \right)\left( r+1 \right)!$ and $\left( n-r-1 \right)!$ can be written as $\left( n-r-1 \right)\left( n-r-2 \right)!$
So we get –
$\dfrac{a}{b}=\dfrac{\left( r+2 \right)\left( r+1 \right)!\left( n-r-2 \right)!}{\left( r+1 \right)!\left( n-r-1 \right)\left( n-r-2 \right)!}$
By cancelling the common factors from numerator and denominator, we get –
$\dfrac{b}{c}=\dfrac{r+2}{n-r-1}$
By cross multiplication, we get –
$\begin{align}
  & b\left( n-r-1 \right)=c\left( r+2 \right) \\
 & bn-br-b=cr+2c \\
\end{align}$
By subtracting ‘$2c$ ’ and adding ‘$br$’ on both sides, we get –
$\begin{align}
  & bn-b-2c=cr+br \\
 & bn-b-2c=r\left( b+c \right) \\
\end{align}$
By dividing $b+c$ on the both sides, we get –
$r=\dfrac{bn-b-2c}{b+c}$ …………………….. (2)
By equalizing the values of ‘r’ from equation (1) and (2) to get the value of ‘n’, we get –
$\dfrac{an-b}{a+b}=\dfrac{bn-b-2c}{b+c}$
By cross multiplication we get –
$\left( an-b \right)\left( b+c \right)=\left( bn-b-2c \right)\left( a+b \right)$
By simplifying we get –
$abn+acn-{{b}^{2}}-bc=abn+{{b}^{2}}n-ab-{{b}^{2}}-2ac-2bc$
By cancelling and transferring the above equation, we get –
$ab+2ac+2bc-bc=-acn+{{b}^{2}}n$
By taking ‘n’ as common at one side, we get –
$ab+2a+bc=n\left( {{b}^{2}}-ac \right)$
By dividing $ac-{{b}^{2}}$on both sides, we get –
$n=\dfrac{bc+ab+2ac}{{{b}^{2}}-ac}$
Hence, option (A) is the correct answer.

Note: Generally students can make mistakes while solving this question. They may take $x,{{x}^{2}},{{x}^{3}}$ as the three consecutive coefficient terms or can take directly the consecutive coefficients $\left( r,r+1,r+2 \right)$ as a, b, c, which may lead to get them a wrong answer.