Question

If the coefficients of $ {\text{}}{{\text{r}}^{{\text{th}}}} $ , $ {{\text{(r + 1)}}^{{\text{th}}}} $ and $ {{\text{(r + 2)}}^{{\text{th}}}} $ terms in the binomial expansion of $ {\left( {1 + x} \right)^n} $ are in A.P., then show that $ {n^2} - (4r + 1)n + 4{r^2} - 2 = 0 $ .

Answer 1

Hint: Write down the general form of $ {\text{}}{{\text{r}}^{{\text{th}}}} $ term of binomial expansion of $ {\left( {1 + x} \right)^n} $. For a set of values in arithmetic progression, the sum of the first and third term in the set is equal to twice of the second term.

Formula used:
$\Rightarrow$ The $ {n^{th}} $ term of the binomial expansion of $ {(1 + x)^n} $ is: $ {}^n{C_r}{x^r} $ .
$\Rightarrow$ For three terms $ {a_n},{a_{n + 1}},{a_{n + 2}} $ in arithmetic progression, $ 2{a_{n + 1}} = {a_n} + {a_{n + 2}} $ .
$\Rightarrow$ $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $

Complete step by step solution:
We’ve been given that the $ {\text{}}{{\text{r}}^{{\text{th}}}} $ , $ {{\text{(r + 1)}}^{{\text{th}}}} $ and $ {{\text{(r + 2)}}^{{\text{th}}}} $ terms of the binomial expression of $ {\left( {1 + x} \right)^n} $ are in arithmetic progression. Let’s start by finding these terms in the binomial expression. We know that the $ {\text{}}{{\text{r}}^{{\text{th}}}} $ term of the binomial expression can be written as:
$ \Rightarrow {\text{}}{{\text{r}}^{{\text{th}}}} $ term: $ {}^n{C_{r - 1}}{(1)^{n - r + 1}}{x^{r - 1}} = {}^n{C_{r - 1}}{x^{r + 1}} $
Similarly, we can write
$ \Rightarrow {{\text{(r + 1)}}^{{\text{th}}}} $ term $ = {}^n{C_r}{x^r} $ and
$ \Rightarrow {{\text{(r + 2)}}^{{\text{th}}}} $ term $ = {}^n{C_{r + 1}}{x^{r + 1}} $ .
Now since the coefficients of these three terms are in arithmetic progression, we can write
 $ \Rightarrow 2{}^n{C_r} = {}^n{C_{r - 1}} + {}^n{C_{r + 1}} $
Dividing both sides by $ {}^n{C_r} $ , we get
$ \Rightarrow 2 = \dfrac{{{}^n{C_{r - 1}}}}{{{}^n{C_r}}} + \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} $
Since $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ , we can write
$ \Rightarrow 2 = \dfrac{{\dfrac{{n!}}{{(r - 1)!(n - (r - 1))!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}} + \dfrac{{\dfrac{{n!}}{{(r + 1)!(n - (r + 1))!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}} $
Using the property that $ r! = r \times (r - 1)! $ , we can simplify the above expression as
 $ \Rightarrow 2 = \dfrac{r}{{n - r + 1}} + \dfrac{{n - r}}{{r + 1}} $
Taking the LCM of the two terms on the right hand side of the equation, we get:
 $ \Rightarrow 2 = \dfrac{{r(r + 1) + (n - r)(n - r + 1)}}{{(n - r + 1)(r + 1)}} $
On simplifying the numerator, we get:
 $ \Rightarrow 2 = \dfrac{{({r^2} + r) + ({n^2} - 2nr + {r^2} + n - r)}}{{(n - r + 1)(r + 1)}} $
On multiplying both sides by $ (n - r + 1)(r + 1) $ , we get
 $ \Rightarrow 2(n - r + 1)(r + 1) = ({r^2} + r) + ({n^2} - 2nr + {r^2} + n - r) $
which can be further simplified to
 $ \Rightarrow {n^2} - (4r + 1)n + 4{r^2} - 2 = 0 $
Hence, we have proved that if the coefficients of $ {\text{}}{{\text{r}}^{{\text{th}}}} $ , $ {{\text{(r + 1)}}^{{\text{th}}}} $ and $ {{\text{(r + 2)}}^{{\text{th}}}} $ terms in the binomial expansion of $ {\left( {1 + x} \right)^n} $ are in A.P. , $ {n^2} - (4r + 1)n + 4{r^2} - 2 = 0 $ .

Note:
Since the terms of the binomial expansion are in A.P., we must realize to use the property $ 2{a_{n + 1}} = {a_n} + {a_{n + 2}} $ which helps us in relating the coefficients of the binomial expansion. Certain identities of permutation/combination operations can also help in simplifying the solution for e.g. $ \dfrac{{{}^n{C_{r - 1}}}}{{{}^n{C_r}}} = \dfrac{r}{{n - r + 1}} $ .