Question

In the expansion of ${\left( {{\text{1 + x}}} \right)^{\text{n}}}$, the coefficient of ${14^{{\text{th}}}}$, ${\text{1}}{{\text{5}}^{{\text{th}}}}$ and ${\text{1}}{{\text{6}}^{{\text{th}}}}$ terms are in Arithmetic Progression. Find n.

Answer 1

Hint: To find n, we use binomial expansion. Then we get the coefficients of ${14^{{\text{th}}}}$, ${\text{1}}{{\text{5}}^{{\text{th}}}}$and ${\text{1}}{{\text{6}}^{{\text{th}}}}$ terms. We use the relation [If a, b, c are in A.P then 2b = a + c].

Complete Step-by-Step solution:
As we know, the binomial expansion of ${\left( {{\text{a + b}}} \right)^{\text{n}}}$is

${\left( {{\text{a + b}}} \right)^{\text{n}}}$ = ${}^{\text{n}}{{\text{C}}_0}{{\text{a}}^{\text{n}}} + {}^{\text{n}}{{\text{C}}_1}{{\text{a}}^{{\text{n - 1}}}}{{\text{b}}^1} + {}^{\text{n}}{{\text{C}}_2}{{\text{a}}^{{\text{n - 2}}}}{{\text{b}}^2} + .....{}^{\text{n}}{{\text{C}}_{\text{r}}}{{\text{a}}^{{\text{n - r}}}}{{\text{b}}^{\text{r}}} + ... + {}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{b}}^{\text{n}}}$

Comparing this to${\left( {{\text{1 + x}}} \right)^{\text{n}}}$, we get a = 1 and b = x, the equation becomes
${\left( {{\text{1 + x}}} \right)^{\text{n}}}$ = ${}^{\text{n}}{{\text{C}}_0}{{\text{1}}^{\text{n}}} + {}^{\text{n}}{{\text{C}}_1}{{\text{1}}^{{\text{n - 1}}}}{{\text{x}}^1} + {}^{\text{n}}{{\text{C}}_2}{{\text{1}}^{{\text{n - 2}}}}{{\text{x}}^2} + .... + {}^{\text{n}}{{\text{C}}_{\text{r}}}{{\text{1}}^{{\text{n - r}}}}{{\text{x}}^{\text{r}}} + ... + {}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}$

Coefficients of${14^{{\text{th}}}}$, ${\text{1}}{{\text{5}}^{{\text{th}}}}$ and ${\text{1}}{{\text{6}}^{{\text{th}}}}$terms are ${}^{\text{n}}{{\text{C}}_{{\text{13}}}},{\text{ }}{}^{\text{n}}{{\text{C}}_{{\text{14}}}}{\text{ and }}{}^{\text{n}}{{\text{C}}_{{\text{15}}}}$respectively.​

And they are in A.P, so we apply the formula 2b = a + c, where a, b and c are ${}^{\text{n}}{{\text{C}}_{{\text{13}}}},{\text{ }}{}^{\text{n}}{{\text{C}}_{{\text{14}}}}{\text{ and }}{}^{\text{n}}{{\text{C}}_{{\text{15}}}}$ respectively. We get,

⟹${\text{2}}{}^{\text{n}}{{\text{C}}_{14}} = {}^{\text{n}}{{\text{C}}_{13}} + {}^{\text{n}}{{\text{C}}_{15}}$
We know${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$, using this we get
$
   \Rightarrow \dfrac{{{\text{2n}}!}}{{14!\left( {{\text{n}} - 14} \right)!}} = \dfrac{{{\text{n}}!}}{{13!\left( {{\text{n}} - 13} \right)!}} + \dfrac{{{\text{n}}!}}{{15!\left( {{\text{n}} - 15} \right)!}} \\
   \Rightarrow \dfrac{{\text{2}}}{{14 \times 13!\left( {{\text{n - 14}}} \right)\left( {{\text{n}} - 15} \right)!}} = \dfrac{1}{{13!\left( {{\text{n}} - 13} \right)\left( {{\text{n - 14}}} \right)\left( {{\text{n - 15}}} \right)!}} + \dfrac{1}{{15 \times 14 \times 13!\left( {{\text{n}} - 15} \right)!}} \\
$
[We converted all the denominators to (n-15)! and 13! to simplify]
$
   \Rightarrow \dfrac{2}{{14\left( {{\text{n - 14}}} \right)}} = \dfrac{1}{{\left( {{\text{n - 14}}} \right)\left( {{\text{n - 13}}} \right)}} + \dfrac{1}{{14 \times 15}} \\
   \Rightarrow \dfrac{1}{{{\text{n - 14}}}}\left[ {\dfrac{1}{7} - \dfrac{1}{{{\text{n - 13}}}}} \right] = \dfrac{1}{{210}} \\
   \Rightarrow \dfrac{1}{{{\text{n - 14}}}}\left[ {\dfrac{{{\text{n - 13 - 7}}}}{{7{\text{n - 91}}}}} \right] = \dfrac{1}{{210}} \\
   \Rightarrow \dfrac{{{\text{n - 20}}}}{{{\text{7}}{{\text{n}}^2} - 91{\text{n - 98n + 1274}}}} = \dfrac{1}{{210}} \\
$
$
   \Rightarrow 210{\text{n - 4200 = 7}}{{\text{n}}^2} - 189{\text{n + 1274}} \\
   \Rightarrow {\text{7}}{{\text{n}}^2} - 399{\text{n + 5474 = 0}} \\
   \Rightarrow {{\text{n}}^2} - 57{\text{n + 782 = 0}} \\
   \Rightarrow {{\text{n}}^2} - 34{\text{n - 23n + 782 = 0}} \\
   \Rightarrow {\text{n}}\left( {{\text{n - 34}}} \right) - 23\left( {{\text{n - 34}}} \right) = 0 \\
   \Rightarrow \left( {{\text{n - 23}}} \right)\left( {{\text{n - 34}}} \right) = 0 \\
   \Rightarrow {\text{n = 23 or 34}} \\
$

Hence n = 23 or 34.

Note: The key in solving such types of problems is having adequate knowledge in binomial expansion and formulae of Arithmetic Progression. Using the formula of the middle term of an AP to equate the given three consecutive terms is the vital step of solving this problem. We then simplify the equation and determine the value of n.