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Question: If the second, third and fourth terms in the expansion of \({\left( {x + a} \right)^n}\) are 240, 72...

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

Explanation

Solution

Hint- Here we will proceed by using the concept of binomial expansion that is (x+y)n=k=0n(kn)xnkyk=k=0n(kn)xkynk{\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 (a+b)n{\left( {a + b} \right)^n} is
Tr+1=nCr(a)n+r(b)r{T^{r + 1}} = {}^n{C_r}{\left( a \right)^{n + r}}{\left( b \right)^r}….. (1)
Given that second term of (x+b)n{\left( {x + b} \right)^n} is 240
That is, T2=240 T1+1=240  {T_2} = 240 \\\ {T_{1 + 1}} = 240 \\\
Putting r=1r = 1, a=xa = x and b=ab = a
Tr+1=nCr(x)n1(a)1{T_{r + 1}} = {}^n{C_r}{\left( x \right)^{n - 1}}{\left( a \right)^1}
T2=nC1xn1a\Rightarrow {T_2} = {}^n{C_1}{x^{n - 1}}a
240=nC1xn1a\Rightarrow 240 = {}^n{C_1}{x^{n - 1}}a ….. (2)
Third term of (x+a)n{\left( {x + a} \right)^n} is 720
That is T3=720{T_3} = 720
T2+1=720\Rightarrow {T_{2 + 1}} = 720
Putting r=2r = 2, a=xa = x and b=ab = a in (1)
T2+1=nC2(x)n2.(b)2{T_{2 + 1}} = {}^n{C_2}{\left( x \right)^{n - 2}}.{\left( b \right)^2}
720=nC2xn2a2\Rightarrow 720 = {}^n{C_2}{x^{n - 2}}{a^2} ….. (3)
Fourth term of (x+a)n{\left( {x + a} \right)^n} is 1080
That is T4=1080{T_4} = 1080
T3+1=1080{T_{3 + 1}} = 1080
Putting r=3r = 3, a=xa = x and b=ab = a in (1)
T3+1=nCrxn3a3{T_{3 + 1}} = {}^n{C_r}{x^{n - 3}}{a^3}
1080=nC3xn3a3\Rightarrow 1080 = {}^n{C_3}{x^{n - 3}}{a^3} …. (4)
Also,
Dividing (3)\left( 3 \right)by (2)\left( 2 \right)
720240=nC2xn2a2nC1xn1a\dfrac{{720}}{{240}} = \dfrac{{{}^n{C_2}{x^{n - 2}}{a^2}}}{{{}^n{C_1}{x^{n - 1}}a}}
3=nC2nC1×xn2xn1×a2a 3=n!2!(n2)!n!1!(n1)!×xn2(n1)×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} \\\
3=n!2!(n2)!×1!(n1)!n!×xn2n+1×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
3=n!2(n2)!×1(n1)(n2)!n!×x1×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
3=(n1)2×ax\Rightarrow 3 = \dfrac{{\left( {n - 1} \right)}}{2} \times \dfrac{a}{x}
By cross multiplication, we will get
3×2=(n1)×ax\Rightarrow 3 \times 2 = \left( {n - 1} \right) \times \dfrac{a}{x}
6=(n1)×ax\Rightarrow 6 = \left( {n - 1} \right) \times \dfrac{a}{x}
6n1=ax\Rightarrow \dfrac{6}{{n - 1}} = \dfrac{a}{x} ….. (A)
Now dividing (4) and (3)
1080720=nC3xn3a3nC2xn2a2\Rightarrow \dfrac{{1080}}{{720}} = \dfrac{{{}^n{C_3}{x^{n - 3}}{a^3}}}{{{}^n{C_2}{x^{n - 2}}{a^2}}}
32=nC3nC2×xn3xn2×a3a2\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}}}
32=n!3!(n3)!n!2(n2)!×xn3n2×a32 32=n!3!(n3)!×2!(n2)!n!×x1×a1 \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} \\\
32=n!3(2)!(n3)!×2!(n2)(n3)!n!×ax 32=(n2)3×ax  \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} \\\
92(n2)=ax\Rightarrow \dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{a}{x} …. (B)
Now our equations are
6n1=ax\dfrac{6}{{n - 1}} = \dfrac{a}{x} …. (A)
92(n2)=ax\dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{a}{x} ….. (B)
Equating (A) and (B)
92(n2)=6n1\Rightarrow \dfrac{9}{{2\left( {n - 2} \right)}} = \dfrac{6}{{n - 1}}
By cross multiply, we will get
9×(n1)=6×2(n2)\Rightarrow 9 \times \left( {n - 1} \right) = 6 \times 2\left( {n - 2} \right)
9n9=12n24\Rightarrow 9n - 9 = 12n - 24
15=3n 153=n 5=n n=5  \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 (a+b)m{\left( {a + b} \right)^m} of a binomial (a+b)\left( {a + b} \right) as a certain sum of products aibi{a^i}{b^i}, such as (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab. By this method we can easily solve this question.