Question

If the second, third and fourth terms in the expansion of ${\left( {x + a} \right)^n}$ are 240, 720, 1080 respectively, then the value of n is
A) 15
B) 20
C) 10
D) 5

Answer 1

Hint- Here we will proceed by using the concept of binomial expansion that is ${\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( {_k^n} \right)} {x^{n - k}}{y^k} = \sum\limits_{k = 0}^n {\left( {_k^n} \right){x^k}} {y^{n - k}}$.

Complete step-by-step answer:
We know that general term of ${\left( {a + b} \right)^n}$ is
${T^{r + 1}} = {}^n{C_r}{\left( a \right)^{n + r}}{\left( b \right)^r}$….. (1)
Given that second term of ${\left( {x + b} \right)^n}$ is 240
 That is, $
  {T_2} = 240 \\
  {T_{1 + 1}} = 240 \\
 $
  Putting $r = 1$, $a = x$ and $b = a$
  ${T_{r + 1}} = {}^n{C_r}{\left( x \right)^{n - 1}}{\left( a \right)^1}$
  $\Rightarrow {T_2} = {}^n{C_1}{x^{n - 1}}a$
  $\Rightarrow 240 = {}^n{C_1}{x^{n - 1}}a$ ….. (2)
 Third term of ${\left( {x + a} \right)^n}$ is 720
 That is ${T_3} = 720$
$\Rightarrow {T_{2 + 1}} = 720$
Putting $r = 2$, $a = x$ and $b = a$ in (1)
 ${T_{2 + 1}} = {}^n{C_2}{\left( x \right)^{n - 2}}.{\left( b \right)^2}$
$\Rightarrow 720 = {}^n{C_2}{x^{n - 2}}{a^2}$ ….. (3)
Fourth term of ${\left( {x + a} \right)^n}$ is 1080
That is ${T_4} = 1080$
${T_{3 + 1}} = 1080$
Putting $r = 3$, $a = x$ and $b = a$ in (1)
${T_{3 + 1}} = {}^n{C_r}{x^{n - 3}}{a^3}$
$\Rightarrow 1080 = {}^n{C_3}{x^{n - 3}}{a^3}$ …. (4)
Also,
Dividing $\left( 3 \right)$by $\left( 2 \right)$
$\dfrac{{720}}{{240}} = \dfrac{{{}^n{C_2}{x^{n - 2}}{a^2}}}{{{}^n{C_1}{x^{n - 1}}a}}$
$
 \Rightarrow 3 = \dfrac{{{}^n{C_2}}}{{{}^n{C_1}}} \times \dfrac{{{x^{n - 2}}}}{{{x^{n - 1}}}} \times \dfrac{{{a^2}}}{a} \\
 \Rightarrow 3 = \dfrac{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}{{\dfrac{{n!}}{{1!\left( {n - 1} \right)!}}}} \times {x^{n - 2 - \left( {n - 1} \right) }\times a} \\
 $
$\Rightarrow 3 = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} \times \dfrac{{1!\left( {n - 1} \right)!}}{{n!}} \times {x^{n - 2 - n + 1}} \times a$
$\Rightarrow 3 = \dfrac{{n!}}{{2\left( {n - 2} \right)!}} \times \dfrac{{1\left( {n - 1} \right)\left( {n - 2} \right)!}}{{n!}} \times {x^{ - 1}} \times a$
$\Rightarrow 3 = \dfrac{{\left( {n - 1} \right)}}{2} \times \dfrac{a}{x}$
By cross multiplication, we will get
$\Rightarrow 3 \times 2 = \left( {n - 1} \right) \times \dfrac{a}{x}$
$\Rightarrow 6 = \left( {n - 1} \right) \times \dfrac{a}{x}$
$\Rightarrow \dfrac{6}{{n - 1}} = \dfrac{a}{x}$ ….. (A)
Now dividing (4) and (3)
$\Rightarrow \dfrac{{1080}}{{720}} = \dfrac{{{}^n{C_3}{x^{n - 3}}{a^3}}}{{{}^n{C_2}{x^{n - 2}}{a^2}}}$
$\Rightarrow \dfrac{3}{2} = \dfrac{{{}^n{C_3}}}{{{}^n{C_2}}} \times \dfrac{{{x^{n - 3}}}}{{{x^{n - 2}}}} \times \dfrac{{{a^3}}}{{{a^2}}}$
$ \Rightarrow \dfrac{3}{2} = \dfrac{\dfrac{n!}{3!\left( {n - 3} \right)!}}{{\dfrac{{n!}}{{2\left( {n - 2} \right)!}}}} \times {x^{{n - 3} - {n - 2}}} \times {a^{3 - 2}} \\
 \Rightarrow \dfrac{3}{2} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} \times \dfrac{{2!\left( {n - 2} \right)!}}{{n!}} \times {x^{ - 1}} \times {a^1} \\
 $
$
 \Rightarrow \dfrac{3}{2} = \dfrac{{n!}}{{3\left( 2 \right)!\left( {n - 3} \right)!}} \times \dfrac{{2!\left( {n - 2} \right)\left( {n - 3} \right)!}}{{n!}} \times \dfrac{a}{x} \\
  \Rightarrow \dfrac{3}{2} = \dfrac{{\left( {n - 2} \right)}}{3} \times \dfrac{a}{x} \\
 $
$\Rightarrow \dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{a}{x}$ …. (B)
Now our equations are
$\dfrac{6}{{n - 1}} = \dfrac{a}{x}$ …. (A)
$\dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{a}{x}$ ….. (B)
Equating (A) and (B)
$\Rightarrow \dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{6}{{n - 1}}$
By cross multiply, we will get
$\Rightarrow 9 \times \left( {n - 1} \right) = 6 \times 2\left( {n - 2} \right)$
$\Rightarrow 9n - 9 = 12n - 24$
$
  \Rightarrow 15 = 3n \\
  \Rightarrow \dfrac{{15}}{3} = n \\
 \Rightarrow 5 = n \\
 \Rightarrow n = 5 \\
 $
Hence, option D is the correct answer.

Note- In this question it should be noted that we use the basics of binomial expansion, elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. Also this theorem specifies the expansion of any power ${\left( {a + b} \right)^m}$ of a binomial $\left( {a + b} \right)$ as a certain sum of products ${a^i}{b^i}$, such as ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. By this method we can easily solve this question.