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Question: If \({a_1},{a_2},{a_3},{a_4}\) are the coefficients of any four consecutive term in the expansion of...

If a1,a2,a3,a4{a_1},{a_2},{a_3},{a_4} are the coefficients of any four consecutive term in the expansion of (1+x)n{\left( {1 + x} \right)^n} , then a1a1+a2+a3a3+a4\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}} is equals to
A - 2a2a2+a3\dfrac{{2{a_2}}}{{{a_2} + {a_3}}}
B - −2a2a2+a3\dfrac{{ - 2{a_2}}}{{{a_2} + {a_3}}}
C - 2a2a1+a3\dfrac{{2{a_2}}}{{{a_1} + {a_3}}}
D - −2a2a1+a3\dfrac{{ - 2{a_2}}}{{{a_1} + {a_3}}}

Explanation

Solution

First let us suppose that the four consecutive terms in the expansion is nCr{}^n{C_r} , nCr+1{}^n{C_{r + 1}} , nCr+2{}^n{C_{r + 2}} , nCr+3{}^n{C_{r + 3}} now put these values in a1a1+a2+a3a3+a4\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}} and solve it we get some value after that go through the option and put the value of a1,a2,a3,a4{a_1},{a_2},{a_3},{a_4} that we suppose and in which case both answer were match that is correct answer

Complete step-by-step answer:
In this question it is given that the coefficients of any four consecutive term is a1,a2,a3,a4{a_1},{a_2},{a_3},{a_4}
Let the four consecutive terms in the expansion is nCr{}^n{C_r} , nCr+1{}^n{C_{r + 1}} , nCr+2{}^n{C_{r + 2}} , nCr+3{}^n{C_{r + 3}} i.e. a1,a2,a3,a4{a_1},{a_2},{a_3},{a_4}
where n is the number of term and r is the term in between the expansion ,
Now for the a1a1+a2+a3a3+a4\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}}
Put the value of the a1,a2,a3,a4{a_1},{a_2},{a_3},{a_4} in above equation that we are suppose above ,
nCrnCr+nCr+1+nCr+2nCr+2+nCr+3\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}} + \dfrac{{{}^n{C_{r + 2}}}}{{{}^n{C_{r + 2}} + {}^n{C_{r + 3}}}}
We know that the nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}} on using this on above we get ,
⇒nCrn+1Cr+1+nCr+2n+1Cr+3\Rightarrow \dfrac{{{}^n{C_r}}}{{{}^{n + 1}{C_{r + 1}}}} + \dfrac{{{}^n{C_{r + 2}}}}{{{}^{n + 1}{C_{r + 3}}}}
As we know that the expansion of nCr{}^n{C_r} = n!r!(n−r)!\dfrac{{n!}}{{r!(n - r)!}} , on expanding
⇒n!r!(n−r)!(n+1)!(r+1)!(n−r)!+n!(r+2)!(n−r−2)!(n+1)!(r+3)!(n−r−2)!\Rightarrow \dfrac{{\dfrac{{n!}}{{r!(n - r)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 1)!(n - r )!}}}} + \dfrac{{\dfrac{{n!}}{{(r + 2)!(n - r - 2)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 3)!(n - r - 2)!}}}}
As n!,r!,(n−r)!n!,r!,(n - r)! is common in both numerator and denominator cancel out both we get
On further solving we get ,
⇒1(n+1)(r+1)+1(n+1)(r+3)\Rightarrow \dfrac{1}{{\dfrac{{(n + 1)}}{{(r + 1)}}}} + \dfrac{1}{{\dfrac{{(n + 1)}}{{(r + 3)}}}}
or we can write as
⇒r+1n+1+r+3n+1\Rightarrow \dfrac{{r + 1}}{{n + 1}} + \dfrac{{r + 3}}{{n + 1}}
⇒2(r+2)n+1\Rightarrow \dfrac{{2(r + 2)}}{{n + 1}}
Now we have to check from the option as in the option (A) 2a2a2+a3\dfrac{{2{a_2}}}{{{a_2} + {a_3}}}
Put the value of a2,a3{a_2},{a_3} as nCr+1{}^n{C_{r + 1}} nCr+2{}^n{C_{r + 2}}
⇒2nCr+1nCr+1+nCr+2\Rightarrow \dfrac{{2{}^n{C_{r + 1}}}}{{{}^n{C_{r + 1}} + {}^n{C_{r + 2}}}} = 2nCr+1n+1Cr+2\dfrac{{2{}^n{C_{r + 1}}}}{{{}^{n + 1}{C_{r + 2}}}} by using the property nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}
Now on expanding we get ,
⇒2n!(r+1)!(n−r−1)!(n+1)!(r+2)!(n−r−1)!\Rightarrow \dfrac{{2\dfrac{{n!}}{{(r+1)!(n - r-1)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 2)!(n - r - 1)!}}}}
As n!,r!,(n−r)!n!,r!,(n - r)! is common in both numerator and denominator cancel out both we get
⇒2(r+2)n+1\Rightarrow \dfrac{{2(r + 2)}}{{n + 1}}
Hence from above checking option (A) is the correct answer.

Note: The formula for finding the term in which x−m{x^{ - m}} come as Tr+1=nα−mα+β{T_{r + 1}} = \dfrac{{n\alpha - m}}{{\alpha + \beta }} where m equals to the power of x . for expression (xα+1xβ)n{\left( {{x^\alpha } + \dfrac{1}{{{x^\beta }}}} \right)^n}.
Always expand the factorial at the end of the solution otherwise the calculation becomes difficult.