Question

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive term in the expansion of ${\left( {1 + x} \right)^n}$ , then $\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}}$ is equals to
A - $\dfrac{{2{a_2}}}{{{a_2} + {a_3}}}$
B - $\dfrac{{ - 2{a_2}}}{{{a_2} + {a_3}}}$
C - $\dfrac{{2{a_2}}}{{{a_1} + {a_3}}}$
D - $\dfrac{{ - 2{a_2}}}{{{a_1} + {a_3}}}$

Answer 1

Hint: First let us suppose that the four consecutive terms in the expansion is ${}^n{C_r}$ , ${}^n{C_{r + 1}}$ , ${}^n{C_{r + 2}}$ , ${}^n{C_{r + 3}}$ now put these values in $\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}}$ and solve it we get some value after that go through the option and put the value of ${a_1},{a_2},{a_3},{a_4}$ 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 ${a_1},{a_2},{a_3},{a_4}$
Let the four consecutive terms in the expansion is ${}^n{C_r}$ , ${}^n{C_{r + 1}}$ , ${}^n{C_{r + 2}}$ , ${}^n{C_{r + 3}}$ i.e. ${a_1},{a_2},{a_3},{a_4}$
where n is the number of term and r is the term in between the expansion ,
Now for the $\dfrac{{{a_1}}}{{{a_1} + {a_2}}} + \dfrac{{{a_3}}}{{{a_3} + {a_4}}}$
Put the value of the ${a_1},{a_2},{a_3},{a_4}$ in above equation that we are suppose above ,
$\dfrac{{{}^n{C_r}}}{{{}^n{C_r} + {}^n{C_{r + 1}}}} + \dfrac{{{}^n{C_{r + 2}}}}{{{}^n{C_{r + 2}} + {}^n{C_{r + 3}}}}$
We know that the ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$ on using this on above we get ,
$\Rightarrow \dfrac{{{}^n{C_r}}}{{{}^{n + 1}{C_{r + 1}}}} + \dfrac{{{}^n{C_{r + 2}}}}{{{}^{n + 1}{C_{r + 3}}}}$
As we know that the expansion of ${}^n{C_r}$ = $\dfrac{{n!}}{{r!(n - r)!}}$ , on expanding
\[\Rightarrow \dfrac{{\dfrac{{n!}}{{r!(n - r)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 1)!(n - r )!}}}} + \dfrac{{\dfrac{{n!}}{{(r + 2)!(n - r - 2)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 3)!(n - r - 2)!}}}}\]
As $n!,r!,(n - r)!$ is common in both numerator and denominator cancel out both we get
On further solving we get ,
\[\Rightarrow \dfrac{1}{{\dfrac{{(n + 1)}}{{(r + 1)}}}} + \dfrac{1}{{\dfrac{{(n + 1)}}{{(r + 3)}}}}\]
or we can write as
$\Rightarrow \dfrac{{r + 1}}{{n + 1}} + \dfrac{{r + 3}}{{n + 1}}$
$\Rightarrow \dfrac{{2(r + 2)}}{{n + 1}}$
 Now we have to check from the option as in the option (A) $\dfrac{{2{a_2}}}{{{a_2} + {a_3}}}$
Put the value of ${a_2},{a_3}$ as ${}^n{C_{r + 1}}$ ${}^n{C_{r + 2}}$
$\Rightarrow \dfrac{{2{}^n{C_{r + 1}}}}{{{}^n{C_{r + 1}} + {}^n{C_{r + 2}}}}$ = $\dfrac{{2{}^n{C_{r + 1}}}}{{{}^{n + 1}{C_{r + 2}}}}$ by using the property ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$
Now on expanding we get ,
$ \Rightarrow \dfrac{{2\dfrac{{n!}}{{(r+1)!(n - r-1)!}}}}{{\dfrac{{(n + 1)!}}{{(r + 2)!(n - r - 1)!}}}}$
As $n!,r!,(n - r)!$ is common in both numerator and denominator cancel out both we get
 $ \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 ${x^{ - m}}$ come as ${T_{r + 1}} = \dfrac{{n\alpha - m}}{{\alpha + \beta }}$ where m equals to the power of x . for expression ${\left( {{x^\alpha } + \dfrac{1}{{{x^\beta }}}} \right)^n}$.
 Always expand the factorial at the end of the solution otherwise the calculation becomes difficult.