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Question: The second, third and fourth terms in the binomial expression \[{({\text{x}} + {\text{a}})^{\text{n}...

The second, third and fourth terms in the binomial expression (x+a)n{({\text{x}} + {\text{a}})^{\text{n}}} are 240,720,1080240,720,1080 respectively. Find the value of x, a and n{\text{x, a and n}}.

Explanation

Solution

By substituting all the terms in the binomial expression formula we will be getting three equations. By dividing the equations, we will be getting the next two equations. By equating the two equations we will be getting the one value. By substituting the one values in the equations we will be getting all the values.

Useful formula:
The formulae used in this question are,
The formula for binomial expression is,
(x+a)n⇒Tr + 1=nCr(x)n−r(a)r{({\text{x}} + {\text{a}})^{\text{n}}} \Rightarrow {{\text{T}}_{{\text{r + 1}}}} = {{\text{n}}_{{{\text{C}}_{\text{r}}}}}{({\text{x}})^{{\text{n}} - {\text{r}}}}{({\text{a}})^{\text{r}}}
Where,
a{\text{a}} be the first of the expression
r{\text{r}} be the term number
n{\text{n}} be the exponent on the binomial
The formula for n!{\text{n}}! is,
n!=n×(n−1)!{\text{n}}! = {\text{n}} \times ({\text{n}} - 1)!
Where,
n{\text{n}} be the required number
The formula for number of combinations is,
nCr=n!r!(n−r)!{{\text{n}}_{{{\text{C}}_{\text{r}}}}} = \dfrac{{{\text{n!}}}}{{{\text{r!(n}} - {\text{r)!}}}}
Where,
n{\text{n}} be the number of objects in the expression
r{\text{r}} be the number of choosing objects

Complete step by step solution:
The data given in the question are,
The second term in the above expression is 240240
The third term in the above expression is 720720
The fourth term in the above expression is 10801080

To find the value of x, a and n{\text{x, a and n}}in the given (x+a)n{({\text{x}} + {\text{a}})^{\text{n}}} expression,
The second term in the above expression = $$$$240
⇒T2=240\Rightarrow {{\text{T}}_2} = 240
The above can also be written as,
⇒T1 + 1=240\Rightarrow {{\text{T}}_{{\text{1 + 1}}}} = 240
By seeing the above equation, we can say that r=1{\text{r}} = 1, and substituting in the formula we get,
⇒nC1(x)n−1(a)1=240...............(1)\Rightarrow {{\text{n}}_{{{\text{C}}_1}}}{({\text{x}})^{{\text{n}} - 1}}{({\text{a}})^1} = 240...............(1)
The third term in the above expression == 720720
⇒T3=720\Rightarrow {{\text{T}}_3} = 720
The above can also be written as,
⇒T2 + 1=720\Rightarrow {{\text{T}}_{{\text{2 + 1}}}} = 720
By seeing the above equation, we can say that r=2{\text{r}} = 2, and substituting in the formula we get,
⇒nC2(x)n−2(a)2=720...............(2)\Rightarrow {{\text{n}}_{{{\text{C}}_2}}}{({\text{x}})^{{\text{n}} - 2}}{({\text{a}})^2} = 720...............(2)
The fourth term in the above expression == 10801080
⇒T4=1080\Rightarrow {{\text{T}}_4} = 1080
The above can also be written as,
⇒T3 + 1=1080\Rightarrow {{\text{T}}_{{\text{3 + 1}}}} = 1080
By seeing the above equation, we can say that r=3{\text{r}} = 3, and substituting in the formula we get,
⇒nC3(x)n−3(a)3=1080...............(3)\Rightarrow {{\text{n}}_{{{\text{C}}_3}}}{({\text{x}})^{{\text{n}} - 3}}{({\text{a}})^3} = 1080...............(3)
By dividing (2) and (1)(2)\,{\text{and (1)}},
⇒nC2(x)n−2(a)2nC1(x)n−1(a)1=720240\Rightarrow \dfrac{{{{\text{n}}_{{{\text{C}}_2}}}{{({\text{x}})}^{{\text{n}} - 2}}{{({\text{a}})}^2}}}{{{{\text{n}}_{{\text{C}}_1}}{{({\text{x}})}^{{\text{n}} - 1}}{{({\text{a}})}^1}}} = \dfrac{{720}}{{240}}
By using the combination formula, we can get,
⇒n!2!(n−2)! ×(x)n−2−(n - 1)×an!1!(n−1)!=3\Rightarrow \dfrac{{\dfrac{{{\text{n!}}}}{{2!({\text{n}} - {\text{2}})!}}\, \times {{({\text{x}})}^{{\text{n}} - 2 - ({\text{n - 1}})}} \times {\text{a}}}}{{\dfrac{{{\text{n!}}}}{{1!({\text{n}} - 1)!}}}} = 3
By simplifying we get,
⇒n!2(n−2)! ×1!(n−1)!n!×(x)−1×a=3\Rightarrow \dfrac{{{\text{n!}}}}{{2({\text{n}} - {\text{2}})!}}\, \times \dfrac{{{\text{1!(n}} - 1){\text{!}}}}{{{\text{n}}!}} \times {({\text{x}})^{ - {\text{1}}}} \times {\text{a}} = 3
By using n!{\text{n!}} formula we get,
⇒(n−1)(n−2)!2(n−2)!×ax=3\Rightarrow \dfrac{{{\text{(n}} - 1)({\text{n}} - 2){\text{!}}}}{{{\text{2(n}} - 2)!}} \times \dfrac{{\text{a}}}{{\text{x}}} = 3
By cancelling we get,
⇒(n−1)2×ax=3\Rightarrow \dfrac{{{\text{(n}} - 1)}}{{\text{2}}} \times \dfrac{{\text{a}}}{{\text{x}}} = 3
⇒(n−1)×ax=6\Rightarrow {\text{(n}} - 1) \times \dfrac{{\text{a}}}{{\text{x}}} = 6
By simplifying the above we get the below equation,
⇒ax=6(n−1)...............(A)\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{{{\text{(n}} - 1)}}...............({\text{A}})
By dividing (3) and (2)(3)\,{\text{and (2)}},
⇒nC3(x)n−3(a)3nC2(x)n−2(a)2=1080720\Rightarrow \dfrac{{{{\text{n}}_{{\text{C}}_3}}{{({\text{x}})}^{{\text{n}} - 3}}{{({\text{a}})}^3}}}{{{{\text{n}}_{{\text{C}}_2}}{{({\text{x}})}^{{\text{n}} - 2}}{{({\text{a}})}^2}}} = \dfrac{{1080}}{{720}}
By using the combination formula, we can get,
⇒n!3!(n−3)! ×(x)n−3−(n−2)×an!2!(n−2)!=32\Rightarrow \dfrac{{\dfrac{{{\text{n!}}}}{{3!({\text{n}} - 3)!}}\, \times {{({\text{x}})}^{{\text{n}} - 3 - ({\text{n}} - 2)}} \times {\text{a}}}}{{\dfrac{{{\text{n!}}}}{{2!({\text{n}} - 2)!}}}} = \dfrac{3}{2}
By simplifying we get,
⇒n!3!(n−3)! ×2!(n−2)!n!×(x)−1×a=32\Rightarrow \dfrac{{{\text{n!}}}}{{3!({\text{n}} - 3)!}}\, \times \dfrac{{{\text{2!(n}} - 2){\text{!}}}}{{{\text{n}}!}} \times {({\text{x}})^{ - {\text{1}}}} \times {\text{a}} = \dfrac{3}{2}
By using n!{\text{n!}} formula we get,
⇒2!(n−2)(n−3)!3!(n−3)!×ax=32\Rightarrow \dfrac{{{\text{2!(n}} - 2)({\text{n}} - 3){\text{!}}}}{{{\text{3!(n}} - 3)!}} \times \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}
By using n!{\text{n!}} formula we get,
⇒2(n−2)3×2×ax=32\Rightarrow \dfrac{{{\text{2(n}} - 2)}}{{3 \times 2}} \times \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}
⇒ax×(n−2)3=32\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} \times \dfrac{{{\text{(n}} - 2)}}{3} = \dfrac{3}{2}
By simplifying the above we get the below equation,
⇒ax=92(n−2)...............(B)\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{9}{{{\text{2(n}} - 2)}}...............({\text{B}})
By equating (A) and (B)({\text{A}})\,{\text{and (B)}},
⇒6n−1=92(n−2)\Rightarrow \dfrac{6}{{{\text{n}} - 1}} = \dfrac{9}{{{\text{2(n}} - 2)}}
By cancelling we get,
⇒2n−1=32(n−2)\Rightarrow \dfrac{2}{{{\text{n}} - 1}} = \dfrac{3}{{{\text{2(n}} - 2)}}
⇒4n−1=3n−2\Rightarrow \dfrac{4}{{{\text{n}} - 1}} = \dfrac{3}{{{\text{n}} - 2}}
Doing cross multiplication, we get,
⇒4(n−2)=3(n−1)\Rightarrow 4({\text{n}} - {\text{2}}) = 3({\text{n}} - {\text{1}})
By solving we get,
⇒4n−8=3n−3\Rightarrow 4{\text{n}} - 8 = 3{\text{n}} - 3
The value of n{\text{n}} is,
⇒n=5\Rightarrow {\text{n}} = 5
Substitute the value of n{\text{n}} in (A)({\text{A}})
⇒ax=65−1\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{{5 - 1}}
⇒ax=64\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{6}{4}
After solving we get,
⇒ax=32\Rightarrow \dfrac{{\text{a}}}{{\text{x}}} = \dfrac{3}{2}
The value of a{\text{a}} is,
⇒a=32x\Rightarrow {\text{a}} = \dfrac{3}{2}{\text{x}}
Substitute the value of a=32x and n=5{\text{a}} = \dfrac{3}{2}{\text{x}}\,{\text{and n}} = 5 in the equation (1)(1)
⇒5C1(x)5−1(32x)=240\Rightarrow {5_{{{\text{C}}_1}}}{({\text{x}})^{5 - 1}}(\dfrac{3}{2}{\text{x}}) = 240
By solving we get,
⇒5(x)4(32x)=240\Rightarrow 5{({\text{x}})^4}(\dfrac{3}{2}{\text{x}}) = 240
⇒x5=240×25×3\Rightarrow {{\text{x}}^5} = \dfrac{{240 \times 2}}{{5 \times 3}}
By cancelling the above we get,
⇒x5=16×2=25\Rightarrow {{\text{x}}^5} = 16 \times 2 = {2^5}
The value of x{\text{x}} is,
⇒x=2\Rightarrow {\text{x}} = 2
∴\therefore The value of x, n and a{\text{x,}}\,{\text{n and a}} is 2, 5 and 32,\,5\,{\text{and 3}} .

Hence, the values of x, n and a{\text{x,}}\,{\text{n and a}} in (x+a)n{({\text{x}} + {\text{a}})^{\text{n}}}
binomial expression are 2, 5 and 32,\,5\,{\text{and 3}}.

Note:
We can divide the equations in any order, we will be getting the same values. While substituting all the values in the binomial expression term formula one by one we will be getting the required terms. This is a small trick for rechecking.