Question

Three successive coefficients in the expression of ${{\left( 1+x \right)}^{n}}$ are 220, 495 and 792. Find the value of n.

Answer 1

Hint: We try to expand the binomial theorem of ${{\left( 1+x \right)}^{n}}$. Then we find out the general term and its coefficient form. We equate it with the given options. Using the formula of combination, we get two equations of two unknowns n and r. we solve it to get the value of n.

Complete step-by-step solution:
We use the expansion of the binomial theorem of ${{\left( 1+x \right)}^{n}}$.
\[{{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{c}_{r}}}{{x}^{r}}=1+nx+{}^{n}{{c}_{2}}{{x}^{2}}+{}^{n}{{c}_{3}}{{x}^{3}}+{}^{n}{{c}_{4}}{{x}^{4}}+.......+{{x}^{n}}\].
So, in every element in the expansion, the coefficient is in the form of \[{}^{n}{{c}_{r}}\].
Three successive coefficients in the expression of ${{\left( 1+x \right)}^{n}}$ are 220, 495 and 792.
Let’s assume that for some $r\in \left[ 0,r-2 \right]$, \[{}^{n}{{c}_{r}}=220\].
The next two elements are consecutive. So, \[{}^{n}{{c}_{r+1}}=495\], \[{}^{n}{{c}_{r+2}}=792\].
Now we have 3 equations to find the value of r.
We also know that \[\dfrac{{}^{n}{{c}_{r+1}}}{{}^{n}{{c}_{r}}}=\dfrac{n!\times r!\times (n-r)!}{n!\times (r+1)!\times (n-r-1)!}=\dfrac{n-r}{r+1}\]. We use this equation to find the relation between those given conditions.
So, \[\dfrac{{}^{n}{{c}_{r+1}}}{{}^{n}{{c}_{r}}}=\dfrac{n-r}{r+1}=\dfrac{495}{220}\] …(i) [given that \[{}^{n}{{c}_{r+1}}=495\], \[{}^{n}{{c}_{r}}=220\]]
Now in this case we use (r+1) in place of r.
We get \[\dfrac{{}^{n}{{c}_{r+2}}}{{}^{n}{{c}_{r+1}}}=\dfrac{n-r-1}{r+2}=\dfrac{792}{495}\] …(ii) [given that \[{}^{n}{{c}_{r+2}}=792\], \[{}^{n}{{c}_{r+1}}=495\]]
We got two equations of two unknowns n and r.
From the first equation, we find out the value of n and put it in the second equation.
We use cross-multiplication.
\[\begin{align}
  & \dfrac{n-r}{r+1}=\dfrac{495}{220} \\
 & \Rightarrow 220n-220r=495r+495 \\
\end{align}\]
Now we find the value of n using binary operations.
\[\begin{align}
  & 220n-220r=495r+495 \\
 & \Rightarrow 220n=495r+220r+495=715r+495 \\
 & \Rightarrow n=\dfrac{715r+495}{220} \\
\end{align}\]
Putting the value of n in the second equation we get the value of r.
We get the simplified form of the second equation by cross-multiplication.
\[\begin{align}
  & \dfrac{n-r-1}{r+2}=\dfrac{792}{495} \\
 & \Rightarrow n-r-1=\dfrac{792}{495}\left( r+2 \right) \\
 & \Rightarrow n=\dfrac{792}{495}\left( r+2 \right)+r+1 \\
\end{align}\]
We put value of n in the equation to get \[\dfrac{715r+495}{220}=\dfrac{792}{495}\left( r+2 \right)+r+1\].
We multiply 1980 both sides to get
\[1980\left( \dfrac{715r+495}{220} \right)=1980\left[ \dfrac{792}{495}\left( r+2 \right)+\left( r+1 \right) \right]\].
Now we apply normal binary operation \[9\left( 715r+495 \right)=4\times 792\left( r+2 \right)+1980(r+1)\].
Then we solve the linear to find the value of r.
\[\begin{align}
  & 6435r+4455=3168\left( r+2 \right)+1980\left( r+1 \right) \\
 & \Rightarrow 6435r+4455=3168r+6336+1980r+1980 \\
 & \Rightarrow 6435r-1980r-3168r=6336+1980-4455 \\
\end{align}\]
We need to apply binary operations of addition and division to find the value of r.
\[\begin{align}
  & 1287r=3861 \\
 & \Rightarrow r=\dfrac{3861}{1287}=3 \\
\end{align}\]
We got a value of r as 3.
Now we put value of r to get value of n. so, \[n=\dfrac{715r+495}{220}=\dfrac{715\times 3+495}{220}=\dfrac{2640}{220}=12\]. The value of n is 12.

Note: We don’t need to worry about the position of the coefficient in the expansion. It’s independent of that. Also, we need to remember that the value of n and r always has to be an integer. it’s one-way to cross-check our answer. The value of n also has to be greater than or equal to r.