Question

Let $ {\left( {1 + {x^2}} \right)^2}{\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^{n - 4} {{a_k}{x^k}} $ . If $ {a_1},{a_2},{a_3} $ are in AP. Find value of n

Answer 1

Hint: In this question, we will use the binomial formula $ {\left( {x + y} \right)^n} = \sum\limits_{r = 1}^n {{}^n{C_r}} {x^{n - r}}{y^r} $ to simplify the expression in the LHS and also expand the expression in RHS and then compare the terms on LHS and RHS to get the terms $ {a_1},{a_2},{a_3} $ . After this, we will have the property that if these three terms are in AP then $ 2{a_2} = {a_1} + {a_3} $ .

Complete step-by-step answer:
Given:- $ {\left( {1 + {x^2}} \right)^2}{\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^{n - 4} {{a_k}{x^k}} $ …… (1)
Also given that $ {a_1},{a_2},{a_3} $ are in A.P.
Condition for three terms to be in A.P is twice the second term will be equal to the sum of the first and third terms.
  $ \Rightarrow 2{a_2} = {a_1} + {a_3} $
Here, we will compare both the sides of equation (1) in order to obtain a expression for $ {a_1},{a_2},{a_3} $
According to binomial theorem,
  $ {\left( {1 + x} \right)^n} = $ \[^n{C_0}{1^n}{x^0}{ + ^n}{C_1}{1^{n - 1}}{x^1}{ + ^n}{C_2}{1^{n - 2}}{x^2} + ......{ + ^n}{C_{n - 1}}{1^1}{x^{n - 1}}{ + ^n}{C_n}{1^0}{x^n}\]
Where, $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ and $ ^n{C_0} = 1{,^n}{C_1} = n{,^n}{C_{n - 1}} = n{,^n}{C_n} = 1 $
  $ \Rightarrow {\left( {1 + x} \right)^n} = 1 + nx{ + ^n}{C_2}{x^2} + ...... + n{x^{n - 1}} + {x^n} $
Taking LHS of equation (1), we get
  $ {\left( {1 + {x^2}} \right)^2}{\left( {1 + x} \right)^n} = \left( {1 + {x^4} + 2{x^2}} \right)\left( {1 + nx{ + ^n}{C_2}{x^2} + ...... + n{x^{n - 1}} + {x^n}} \right) $
  $ \Rightarrow {\left( {1 + {x^2}} \right)^2}{\left( {1 + x} \right)^n} = 1 + nx + \left( {^n{C_2} + 2} \right){x^2} + \left( {^n{C_3} + 2n} \right){x^3} + ...... $ …… (2)

Taking RHS of equation (1) we get
  $ \sum\limits_{k = 0}^{n - } {{a_k}{x^k} = {a_0}{x^0} + {a_1}{x^1}} + {a_3}{x^2} + .... + {a_k}{x^k} = {a_0} + {a_1}x + {a_2}{x^2} + ..... + {a_k}{x^k} $ …… (3)
Now, equating the LHS and RHS of equation 1 using equation 2 and equation 3, we get: $ 1 + nx + \left( {^n{C_2} + 2} \right){x^2} + \left( {^n{C_3} + 2n} \right){x^3} + ...... = {a_0} + {a_1}x + {a_2}{x^2} + ..... + {a_k}{x^k} $ …… (4)
Now comparing the coefficients of $ x,{x^2} $ and $ {x^3} $ on both the sides of eq (4) in orders to obtain $ {a_1},{a_2},{a_3} $
  $ \Rightarrow {a_1} = n,{a_2}{ = ^n}{C_2} + 2,{a_3}{ = ^n}{C_3} + 2n $
Using $ 2{a_2} = {a_1} + {a_3} $ and $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get
  $ 2\left( {^n{C_2} + 2} \right) = n{ + ^n}{C_3} + 2n \Rightarrow 2\left[ {\dfrac{{n!}}{{2!\left( {n - 2} \right)!}} + 2} \right] = 3n + \dfrac{{n!}}{{3!(n - 3)!}} $
  $ \Rightarrow 2\left[ {\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2 \times \left[ {\left( {n - 2} \right)!} \right]}} + 2} \right] = 3n + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{6 \times \left[ {\left( {n - 3} \right)!} \right]}} $
  $ \Rightarrow 2\left[ {\dfrac{{n\left( {n - 1} \right)}}{2} + 2} \right] = 3n + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} \Rightarrow n\left( {n - 1} \right) + 4 = 3n + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} $
Multiply by 6 on both sides of the above equation, we get
  $ \Rightarrow 6{n^2} - 6n + 24 = 18n + n\left( {{n^2} - 3n + 2} \right) $
  $ \Rightarrow 6{n^2} - 6n + 24 = {n^3} - 3{n^2} + 20n $
On rearranging the terms, we get:
  $ \Rightarrow p(n) = {n^3} - 9{n^2} + 26n - 24 = 0 $ .
  $ p(2) = {2^3} - 9 \times {2^2} + 26 \times 2 - 24 = 60 - 60 = 0 $
  $ p(3) = {3^3} - 9 \times {3^2} + 26 \times 3 - 24 = 105 - 105 = 0 $
  $ p(4) = {4^3} - 9 \times {4^2} + 26 \times 4 - 24 = 168 - 168 = 0 $ .
This cubic polynomial is satisfied when we put n = 2 or n = 3 or n= 4.
This means that these are the roots of the cubic polynomial.
Therefore, n = 2, 3, 4.

Note: In this type of question, the main step is to use the binomial expansion to expand the expression in the LHS and RHS side. You should know that if three numbers a, b and c are in AP then middle terms are known as arithmetic mean between the other two numbers and are given as 2b = a + c. Similarly, if three numbers a, b and c are GP then b is called Geometric mean and is given as $ b = \sqrt {ac} $ .